Magnetic Effects of Electric Current - Class 10 Science - Chapter 12 - Notes, NCERT Solutions & Extra Questions
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Notes - Magnetic Effects of Electric Current | Class 10 NCERT | Science
Magnetic Effects of Electric Current Class 10 Notes: Comprehensive Guide
Understanding the magnetic effects of electric current is a crucial part of the Class 10 science curriculum. This topic not only helps in grasping the intricate links between electricity and magnetism but also sets the stage for more advanced studies in physics. Let’s dive into a detailed examination of this concept.
Basic Concepts
Electric Current and Magnetism
In the previous chapter, we learned about the heating effects of electric current. Now, we explore its magnetic effects. A current-carrying conductor behaves like a magnet, and this can be demonstrated through simple experiments.
Historical Insights
Hans Christian Oersted’s Discovery
In 1820, Hans Christian Oersted discovered that a compass needle deflects when placed near a current-carrying wire, unveiling the relationship between electricity and magnetism. This laid the foundation for electromagnetism, a cornerstone of modern technology.
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Magnetic Field and Field Lines
Understanding Magnetic Fields
A magnetic field exists around a magnet, causing a compass needle to deflect. The area around a magnet where its influence can be felt is called the magnetic field, depicted using magnetic field lines.
Experiment with Iron Filings
Place a bar magnet on a paper, sprinkle iron filings, and tap gently. The filings align along the magnetic field lines, forming distinct patterns. This demonstrates the presence and structure of the magnetic field.
Drawing Magnetic Field Lines with a Compass
Using a small compass, you can trace the field lines around a magnet. The needle’s north pole points in the direction of the magnetic field at any given point.
Magnetic Field Due to a Current-Carrying Conductor
Magnetic Field Around a Straight Conductor
When electric current flows through a straight conductor, a magnetic field is produced around it. The direction of this field can be determined using the Right-Hand Thumb Rule.
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Magnetic Field in Different Configurations
Circular Loop
Bending a straight wire into a circular loop and passing current through it creates a unique magnetic field pattern. The field lines in the center of the loop appear as straight lines.
Solenoid
A solenoid is a coil of wire that, when carrying current, produces a magnetic field similar to a bar magnet. The field inside a solenoid is uniform and parallel.
Magnetic Force on a Conductor
Force on a Current-Carrying Conductor in a Magnetic Field
A current-carrying conductor placed in a magnetic field experiences a force. This phenomenon is described by Fleming’s Left-Hand Rule, aiding the construction of devices such as electric motors.
Practical Applications
Electromagnets
Electromagnets are made by coiling wire around a soft iron core. They are widely used in various applications, from electric bells to MRI machines in medical fields.
Electric Motors
Electric motors convert electrical energy into mechanical energy using the interaction between magnetic fields and current-carrying conductors.
Magnetism in Medicine
Magnetic Resonance Imaging (MRI)
MRI machines leverage the magnetic fields produced by current-carrying wires to create detailed images of the body's internal structures, aiding in medical diagnoses.
Understanding Domestic Electric Circuits
Components of a Domestic Electric Circuit
A typical household electric circuit includes live, neutral, and earth wires, and ensures safe distribution and usage of electric power.
Safety Measures in Domestic Circuits
Implementing proper fuses and earthing in domestic circuits prevents hazards like overloading and short circuits, ensuring the safety of appliances and users.
Conclusion
Understanding the magnetic effects of electric current not only equips you with essential knowledge for your Class 10 exams but also forms a foundation for advanced studies in physics and engineering.
FAQs about Magnetic Effects of Electric Current
What are the magnetic effects of electric current? The magnetic effects of electric current refer to the phenomenon where a current-carrying conductor generates a magnetic field around it.
How does a current-carrying wire behave like a magnet? A current-carrying wire produces a circular magnetic field around it, analogous to a magnet’s field.
What is the Right-Hand Thumb Rule? The Right-Hand Thumb Rule is used to determine the direction of the magnetic field around a straight conductor carrying current.
For more such explanations and practical insights, keep revising and exploring the wonderful world of science!
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Extra Questions - Magnetic Effects of Electric Current | NCERT | Science | Class 10
A circular conducting wire having uniform resistance per unit length is connected to a battery of $4 \mathrm{~V}$. The total resistance of the conductor is $4 \Omega$. Then the net magnetic field at the centre of the conductor is (Ignore magnetic field of wire present in circuit other than that of circular conducting wire)
(A) $\frac{\mu_{0}}{2}$
(B) $\frac{8 \mu_{0}}{3}$
(C) $2 \mu_{0}$
(D) Zero
The correct option is D Zero.
To find the net magnetic field at the center of the circular conducting wire, we'll use the formula for the magnetic field due to a segment of a circular conductor. For a segment of the conductor, the magnetic field at the center can be given as: $$ B = \frac{\mu_{0} I \theta}{4 \pi R} $$ where:
$\mu_{0}$ is the permeability of free space,
$I$ is the current flowing through the segment,
$\theta$ is the angle subtended by the segment at the center,
$R$ is the radius of the circular segment.
Given that the total resistance of the circular wire is 4 Ω and it is connected to a 4 V battery, the current flowing through the circuit can be calculated using Ohm's Law: $$ I = \frac{V}{R} = \frac{4 \text{ V}}{4 \text{ Ω}} = 1 \text{ A} $$
Since the resistance of the wire is uniform and proportional to its length, different segments will have different resistances.
Let the lengths of the two segments be $l_1$ and $l_2$, respectively. The ratios of their lengths and currents are: $$ \frac{l_1}{l_2} = \frac{l_2}{l_1} = 3 $$
Now let's calculate the magnetic fields due to each of these segments:
For the first segment: $$ B_1 = \frac{\mu_{0} I l_1}{4 \pi R} \times \frac{\pi}{2} $$ Given the ratio provided, the length $l_1 = 3 l_2$, so: $$ B_1 = \frac{\mu_{0} (3 I_2)}{4 \pi R} \times \frac{\pi}{2} $$
For the second segment: $$ B_2 = \frac{\mu_{0} I l_2}{4 \pi R} \times 3\frac{\pi}{2} $$
The net magnetic field at the center of the circular conductor will be the difference between these two components: $$ B_{\text{net}} = B_1 - B_2 $$ $$ B_{\text{net}} = \left(\frac{\mu_{0} (3 I_2)}{4 \pi R} \times \frac{\pi}{2}\right) - \left(\frac{\mu_{0} I l_2}{4 \pi R} \times 3 \frac{\pi}{2}\right) $$
Given the uniform resistance and current distribution, the net magnetic field due to the opposing segments essentially cancels out, leading to: $$ B_{\text{net}} = 0 $$
Therefore, the net magnetic field at the center of the circular conductor is zero.
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Which of the following correctly describes the magnetic field near a long straight wire?
(a) The field consists of straight lines perpendicular to the wire.
(b) The field consists of straight lines parallel to the wire.
(c) The field consists of radial lines originating from the wire.
(d) The field consists of concentric circles centred on the wire.
The correct answer is:
(d) The field consists of concentric circles centred on the wire.
The magnetic field around a long straight wire carrying current forms concentric circles centered on the wire. The direction of the magnetic field lines can be determined by the right-hand rule: if you wrap your right hand around the wire with your thumb pointing in the direction of the current, your fingers will curl in the direction of the magnetic field lines.
At the time of short circuit, the current in the circuit
(a) reduces substantially.
(b) does not change.
(c) increases heavily.
(d) vary continuously.
At the time of a short circuit, the current in the circuit increases heavily. This is because a short circuit causes a very low resistance path for the current to flow, which allows a large amount of current to pass through the circuit. According to Ohm's Law $ V = IR $, where $ V $ is the voltage, $ I $ is the current, and$ R $ is the resistance, if the resistance $ R $ becomes extremely small, the current $ I $ becomes very large if the voltage $ V$ is constant.
Therefore, the correct answer is:
(c) increases heavily.
State whether the following statements are true or false.
(a) The field at the centre of a long circular coil carrying current will be parallel straight lines.
(b) A wire with a green insulation is usually the live wire of an electric supply.
(a) True. The magnetic field at the center of a long circular coil carrying current can indeed be considered as nearly parallel straight lines. This is due to the magnetic field lines being concentric circles around the wire, and at the center, these circles appear as parallel straight lines.
(b) False. A wire with green insulation is not usually the live wire of an electric supply. In many electrical systems, particularly those following international standards (IEC wiring color codes), a green-insulated wire, often with a yellow stripe, is used for the earth or ground connection. The live wire is typically brown, red, or black, but the color code can vary by country and type of installation.
List two methods of producing magnetic fields.
Two common methods of producing magnetic fields are:
Electric Currents: When electric current flows through a conductor (e.g., a wire), it generates a magnetic field around the conductor. The direction of the magnetic field can be determined by the right-hand rule. This principle is utilized in electromagnets, solenoids, and electrical motors.
Permanent Magnets: Materials with magnetic properties can produce a magnetic field without the need for electricity. These materials, such as iron, nickel, and cobalt, have domains with aligned magnetic moments that generate a magnetic field. Permanent magnets retain their magnetism over time and can create persistent magnetic fields.
When is the force experienced by a current–carrying conductor placed in a magnetic field largest?
The force experienced by a current-carrying conductor placed in a magnetic field is largest when:
The current flowing through the conductor is maximized.
The strength of the magnetic field ((B)) in which it is placed is at its highest.
The length of the conductor within the magnetic field is maximized.
The conductor is oriented at an angle of 90 degrees ((90^\circ)) to the magnetic field lines.
This relationship is described by the equation: $$ \vec{F} = I\vec{L} \times \vec{B} $$ where:
$\vec{F}$ is the force experienced by the conductor,
$I$ is the current through the conductor,
$\vec{L}$ is the vector representing the length of the conductor in the direction of the current,
$\vec{B}$ is the magnetic field vector.
The magnitude of the force can also be represented as: $$ F = ILB \sin(\theta) $$ where:
$\theta$ is the angle between $\vec{L}$ and $\vec{B}$.
Given this, the force is largest when $\theta = 90^\circ$, which makes $sin(\theta) = 1$, maximizing the value of $F$. This configuration ensures the current-carrying conductor interacts with the magnetic field in the most impactful way, leading to the largest possible force acting upon it.
Imagine that you are sitting in a chamber with your back to one wall. An electron beam, moving horizontally from back wall towards the front wall, is deflected by a strong magnetic field to your right side. What is the direction of magnetic field?
To determine the direction of the magnetic field, we can use the right-hand rule, specifically for the situation described by Fleming’s Left-Hand Rule because it involves a moving charge (an electron) in a magnetic field. However, since electrons are negatively charged, the actual force direction is opposite to what the left-hand rule predicts.
In the scenario described, the electron beam moves from the back wall towards the front wall and is deflected to your right side. If you were to use the left-hand rule but account for the negative charge, your thumb (representing the direction of the force acting on the electron) points to the right, your middle finger (representing the direction of the electron) points towards the front wall, and therefore your index finger (which would point in the direction of the magnetic field acting on the electron) must point upwards.
So, the magnetic field direction is upwards relative to your position in the chamber.
State the rule to determine the direction of a
(i) magnetic field produced around a straight conductor-carrying current,
(ii) force experienced by a current-carrying straight conductor placed in a magnetic field which is perpendicular to it, and
(iii) current induced in a coil due to its rotation in a magnetic field.
To determine the direction of various magnetic phenomena, several rules have been established in physics. Here are the rules for the cases you've asked about:
(i) Direction of magnetic field produced around a straight conductor carrying current
The right-hand thumb rule is used for this purpose. To apply this rule, you hold the conductor (straight wire) in your right hand so that your thumb points in the direction of the current. Then, the curl of your fingers around the conductor shows the direction of the magnetic field lines encircling the conductor. This rule suggests that the magnetic field lines form concentric circles around the wire, with their direction determined by the right-hand curl.
(ii) Direction of force experienced by a current-carrying straight conductor placed in a magnetic field which is perpendicular to it
Here, the left-hand rule (often referred to as Fleming's Left-Hand Rule) is used. Extend the thumb, forefinger, and middle finger of your left hand so that they are all perpendicular to each other. The Forefinger points in the direction of the magnetic field (from North to South). The Center finger points in the direction of the current (from Positive to Negative). The Thumb then points in the direction of the force or motion of the conductor.
(iii) Direction of current induced in a coil due to its rotation in a magnetic field
For this scenario, Fleming's right-hand rule is applicable, which is often used for generators. To use this rule, extend the thumb, forefinger, and middle finger of your right hand so that they are mutually perpendicular to each other. If the forefinger represents the direction of the magnetic field, and the thumb indicates the motion of the conductor (or the direction of the force), then the middle finger will point in the direction of the induced current.
These rules are essential tools in understanding and analyzing magnetic effects related to electric currents and play a significant role in designing and interpreting experiments and devices based on electromagnetic principles.
When does an electric short circuit occur?
An electric short circuit occurs when there is an unintentional path of low resistance that allows a current to flow along an unintended path. This may happen due to several reasons, such as:
Wiring or insulation damage: Over time, wires may become frayed, or insulation may become damaged, allowing current to flow outside its intended path.
Faulty appliances or devices: Faults within electrical devices can create unintended paths for current.
Accidental bridging of wires: This can happen if conductive materials come into contact with wires, bypassing the usual resistance of the circuit.
Overloading of electrical outlets: Plugging too many devices into an outlet can also result in a short circuit.
Short circuits can lead to high currents flowing through the circuit, which may generate heat and potentially cause fires, damage to equipment, or electric shock. This is why circuits are typically protected by fuses or circuit breakers, which are designed to interrupt the flow of electricity in the event of a short circuit.
What is the function of an earth wire? Why is it necessary to earth metallic appliances?
An earth wire is a crucial safety feature in electrical systems that provides a path for excess electrical current to flow safely into the ground. Its primary function is to prevent electrical shocks. When a fault occurs in an appliance, such as a short circuit or a breakdown in insulation, the earth wire provides a low resistance path for the excess current, allowing it to flow into the ground. This action significantly reduces the risk of electric shock to users by ensuring the outer metallic parts of the appliance do not become energized.
Earthing is necessary for metallic appliances because these appliances can conduct electricity. If a fault occurs and the appliance becomes 'live', anyone who touches it could become a path to the ground for the electricity, leading to a potentially lethal shock. Earthing ensures this excess electricity is directed safely into the ground, protecting users from electric shock.
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A vertical conducting ring of radius $R$ falls vertically under gravity in a horizontal magnetic field of magnitude $B_{0}$. The direction of $B_{0}$ is perpendicular to the plane of the ring. When the speed of the ring becomes $v$:
A. Points $C$ and $D$ are at the same potential.
B. Points $A$ and $B$ are at the same potential.
C. Time to gain speed $v$ is more than $\frac{v}{g}$.
D. No current flows in the ring.
The correct options are: A. Points $C$ and $D$ are at the same potential. C. Time to gain speed $v$ is more than $\frac{v}{g}$
Explanation for option A: Since the induced electromotive force (EMF) depends on the change in magnetic flux through the loop, and the motion is purely vertical, Points $C$ and $D$ align horizontally and do not experience any differential magnetic field or motion-related change in flux across them. This means there is no induced EMF between these points, leaving them at the same electrical potential.
Explanation for option C: The presence of a horizontal magnetic field $B_0$ and a vertical velocity $v$ of the ring induces an EMF across points $A$ and $B$ (diametrically opposite on the ring), which is calculated as $Bv\ell$, where $\ell$ is the component of the ring's circumference parallel to $B_0$. Because of this EMF, a current flows which leads to charge accumulation and a Lorentz force opposing the motion (upwards). The opposing force effectively reduces the net downward acceleration of the ring from pure gravity $g$ to something less. Consequently, the ring takes longer to reach the speed $v$ than it would under gravity alone, hence the time required is greater than $\frac{v}{g}$.
State the rule to determine the direction of a
(i) magnetic field produced around a straight conductor carrying current.
(i) Maxwell's Right-Hand Thumb Rule:
This rule states that when a straight conductor carries a current, it generates a magnetic field around it. The shape of this magnetic field is in the form of concentric circles, centered along the axis of the conductor. The direction in which these circular magnetic lines flow depends on the current's direction. Furthermore, the density of the magnetic field is directly proportional to the current's magnitude. According to this rule, if you grip the conductor in your right hand with the thumb pointing in the direction of the current, your fingers will curl in the direction of the magnetic field lines.
For an inverted image, the sign convention of the height of the image is:
In optics, distances measured downward from the principal axis are considered negative. Since an inverted image is oriented downward with respect to the principal axis, the height of the inverted image is labeled as negative.
The charge in coulombs of 1 gram ion of $\mathrm{N}^{3+}$ is (the charge on an electron is $1.602 \times 10^{-19}$ C) (A) $2.894 \times 10^{5}$ C (B) $3.894 \times 10^{5}$ C (C) $2.894 \times 10^{6}$ C (D) none of these
The correct answer is option A: $$ 2.894 \times 10^{5} , \mathrm{C} $$
Explanation:
The charge on one $\mathrm{N}^{3+}$ ion is given by: $$ Q = Z \times e $$ where $Z = 3$ (as it is $\mathrm{N}^{3+}$) and $e = 1.602 \times 10^{-19}$ C (the charge of an electron).
Calculating the charge on one $\mathrm{N}^{3+}$ ion, we get: $$ Q = 3 \times 1.602 \times 10^{-19} , \mathrm{C} = 4.806 \times 10^{-19} , \mathrm{C} $$
However, we need to find the total charge of 1 gram-ion of $\mathrm{N}^{3+}$. The number of ions in 1 mole (Avogadro's number, $N_A$) is $6.02 \times 10^{23}$. So, for 1 mole of $\mathrm{N}^{3+}$ ions: $$ \text{Total charge} = N_A \times Q = 6.02 \times 10^{23} \times 4.806 \times 10^{-19} , \mathrm{C} $$
Calculating this: $$ \text{Total charge} = 2.894 \times 10^{5} , \mathrm{C} $$
Therefore, the charge in coulombs of 1 gram-ion of $\mathrm{N}^{3+}$ is $2.894 \times 10^{5}$ C.
The units of magnetic flux density are:
A) Tesla B) Weber per square meter ($/(\text{metre})^{2}$) C) Weber per ampere squared (Weber/(ampere)^{2}) D) Tesla-meter
The units of magnetic flux density are correctly represented by the following options:
A) Tesla
B) Weber per square meter $(\text{Weber/metre}^2)$
Magnetic flux density is measured in Tesla, which is equivalent to Weber per square meter $(\text{Weber/m}^2)$. This unit is named in honor of the scientist Nikola Tesla.
The figure shows the current flowing in the coil of wire wound around the soft iron horseshoe core. State the polarities developed at the ends A and B.
The polarities developed at the ends of the coil in the given figure are:
End A: South pole
End B: North pole
The determination of these polarities can be made via the Right Hand Rule for coils, which states that if you curl the fingers of your right hand in the direction of the current, your thumb will point towards the north pole. In the image, the current direction at end B is upwards (aligning with your curled fingers), which means end B is the North pole. Conversely, at end A with the downward current, it would be the South pole.
A magnet is placed on a paper along the east-west direction. A compass, away from it, constantly points along the north-south direction. The reason is:
A. The compass is malfunctioning.
B. Magnetic field lines have a single direction.
C. Magnetic field of the magnet doesn't influence the point.
D. The needle is fixed too tight.
The correct answer is C: Magnetic field of the magnet doesn't influence the point.
Magnetic field lines around a magnet originate from the north pole and terminate at the south pole, forming a loop. These lines continue inside the magnet, moving from the south pole back to the north pole, thus completing the loop. When the compass is placed sufficiently far from the magnet, the magnetic field at that location is very weak. This weak field strength implies that the compass is primarily influenced by Earth's magnetic field rather than the magnet's field, thus still pointing in the north-south direction.
A combination of secondary cells has an equivalent EMF of $2 \mathrm{~V}$, internal resistance of $400 \Omega$. Maximum current that can be drawn is:
A) $0.005 \mathrm{~A}$
B) $0.0005 \mathrm{~A}$
C) $0.010 \mathrm{~A}$
D) $0.010 \mathrm{~A}$
The correct answer is A) $0.005 \mathrm{~A}$.
To understand how to derive this, consider the key formula needed to find the current through the circuit:
$$ i = \frac{V}{R} $$
where $V$ is the voltage (or EMF) of the battery and $R$ is the internal resistance. In this problem, you are provided with an EMF of $2 \mathrm{V}$ and an internal resistance of $400 \Omega$.
To draw the maximum current from the battery, you need to minimize the external resistance connected across it, effectively making it $0 \Omega$. This scenario is known as short-circuiting the battery.
Applying the formula:
$$ i = \frac{2 \mathrm{V}}{400 \Omega} = 0.005 \mathrm{A} $$
This calculation confirms that the maximum current that can be drawn from this configuration is $0.005 \mathrm{~A}$. Thus, the correct answer is A) $0.005 \mathrm{~A}$.
Sort the magnetic strength of electromagnet in ascending order from top to bottom when current is varying.
A) Current $=1 \mathrm{~A}$
B) Current $=3 \mathrm{~A}$
C) Current $=6 \mathrm{~A}$
D) Current $=8 \mathrm{~A}$
E) Current $=12 \mathrm{~A}$
An electromagnet generates a magnetic field when electric current flows through it. The strength of the magnetic field in an electromagnet is directly influenced by the magnitude of the current. In scenarios where the same electromagnet is used, increasing the electric current enhances the magnetic field strength.
In the given question, various currents are listed, but the electromagnet remains constant. Thus, the magnetic strength is proportional to the current. The ascending order of the magnetic strength corresponding to the different currents is as follows:
Current $= 1 \mathrm{~A}$ - Weakest magnetic strength
Current $= 3 \mathrm{~A}$
Current $= 6 \mathrm{~A}$
Current $= 8 \mathrm{~A}$
Current $= 12 \mathrm{~A}$ - Strongest magnetic strength
This ordering is based on the principle that higher currents produce stronger magnetic fields in an electromagnet.
Which of the following is true about current-carrying conductors?
A) They behave as magnets always.
B) A coil of the current-carrying conductor behaves like a magnet.
C) They can never behave as a magnet.
D) A current-carrying conductor is always attracted to a magnet.
The correct answer to the question is Option B: "A coil of the current-carrying conductor behaves like a magnet."
When a conductor through which current flows is coiled, it exhibits magnetic properties. It effectively turns into an electromagnet, characterized by the presence of two magnetic poles. This ability to behave like a magnet allows it to interact with other magnets or magnetic materials in typical magnetic ways—its north pole will repel the north pole of another magnet and attract the south pole, and vice versa.
The middle finger in Fleming's right hand rule represents the direction of which quantity?
A) Motion of conductor
B) Induced current
C) Magnetic field
D) Flow of charges
The correct answer is Option B: Induced current.
In Fleming's right-hand rule, different fingers represent different directions:
The thumb shows the direction of motion of the conductor.
The forefinger indicates the direction of the magnetic field.
The middle finger points in the direction of the induced current.
When two mutually perpendicular simple harmonic motions of the same frequency, amplitude, and phase are superimposed:
A. The resulting motion is uniform circular motion.
B. The resulting motion is a linear simple harmonic motion along a straight line inclined equally to the straight lines of motion of the component ones.
C. The resulting motion is an elliptical motion, symmetrical about the lines of motion of the components.
D. The two S.H.M's will cancel each other.
The correct option is B.
The resulting motion is a linear simple harmonic motion along a straight line inclined equally to the straight lines of motion of the component ones.
Given two simple harmonic motions: $$ Y_1 = A_1 \sin(\omega t) \quad \text{and} \quad Y_2 = A_2 \sin \left(\omega t + \frac{\pi}{2}\right) $$ where $\omega$ is the angular frequency, and $A_1 = A_2$ (since amplitudes are the same). The phase of $Y_2$ is shifted by $\frac{\pi}{2}$, suggesting it leads $Y_1$ by a quarter cycle. The resultant displacement vector is: $$ A = \sqrt{A_1^2 + A_2^2} $$ Angle $\theta$ which it makes with the direction of $Y_1$ is: $$ \theta = \tan^{-1}\left(\frac{A_2}{A_1}\right) = \tan^{-1}(1) = \frac{\pi}{4} $$ This angle implies that the motion is inclined equally to the original axes of motions.
The magnetic field inside the solenoid is:
A. Non-uniform
B. Zero
C. Almost the same at all points
D. Can't be said
The correct answer is C. Almost the same at all points.
The magnetic field inside a solenoid is consistent and uniform, making it almost the same at all points. This uniformity in the magnetic field is comparable to that observed in a bar magnet.
An infinitely long, straight conductor $AB$ is fixed, and a current is passed through it. Another movable straight wire CD of finite length and carrying current is held perpendicular to it and released. Neglect the weight of the wire.
Which of the following options correctly describes the movement of rod CD?
A. The rod CD will move upwards parallel to itself.
B. The rod CD will move downward parallel to itself.
C. The rod CD will move upward and turn clockwise at the same time.
D. The rod CD will move upward and turn anticlockwise at the same time.
The correct answer is C. The rod CD will move upward and turn clockwise at the same time.
When analyzing the motion and orientation of the movable conductor CD, we need to consider the effects of the magnetic field due to the current in conductor AB and the force experienced by CD. Conductor AB, being infinitely long and straight, generates a magnetic field around it. When a current flows through CD, which is held perpendicular to AB, it experiences a magnetic force due to the field of AB.
According to the left-hand rule, where the thumb points in the direction of the current and the fingers in the direction of the magnetic field, the palm faces the direction of the force. Since current flows perpendicularly from CD and interacts with the magnetic field from AB, the force on CD will push it upward.
Additionally, because not all parts of CD are equidistant from AB, there is a non-uniform force distribution along CD, causing not just translational motion but also a rotational force. This rotational force, or torque, will cause CD to rotate. Applying the left-hand rule carefully considering this distribution, CD will experience a torque that causes it to turn clockwise.
Therefore, CD will move upward due to the magnetic force and simultaneously turn clockwise due to the torque created by the non-uniform distribution of the magnetic field's effect on the length of CD.
The force experienced by the magnet is due to the magnetic field of the current-carrying conductor.
A) True
B) False
The correct answer is A) True.
The force experienced by the magnet arises directly from the magnetic field created by the current-carrying conductor. This magnetic field's existence and characteristics depend on the magnitude of the current flowing through the conductor and the distance from the conductor to the point of interest. Importantly, the direction of the magnetic field is perpendicular to the conductor. This interaction governs the magnetic force exerted on nearby magnets.
A coil of insulated copper wire is connected to a galvanometer. What would happen if a bar magnet is: (i) Pushed into the coil. (ii) Withdrawn from inside the coil. (iii) Held stationary inside the coil?
(i) Pushing the bar magnet into the coil leads to a change in the magnetic flux linked with the coil. Consequently, the galvanometer will indicate this change by showing a deflection; let's say it deflects to the right.
(ii) Withdrawing the bar magnet from inside the coil also changes the magnetic flux. However, this change is in the opposite direction compared to when the magnet was pushed in. Therefore, the galvanometer will show a deflection in the opposite direction (to the left, as an example).
(iii) When the bar magnet is held stationary inside the coil, there is no change in the magnetic flux linked with the coil. As a result, the galvanometer will show no deflection.
$P$ and $Q$ are two thin circular coils of the same radius and are subjected to the same rate of change of flux. The magnetic field is perpendicular inwards to their plane and varying with time. If coil $P$ is made of copper and $Q$ is made of iron, then the wrong statement is $\left(\rho_{\text{iron}} > \rho_{\text{copper}}\right)$.
A. The emf induced in the two is the same.
B. The induced current in $P$ is more than that in $Q.
C. Resistance of the two coils is different.
D. The induced currents are the same in both coils.
The correct answer is Option D: The induced currents are the same in both coils. This statement is incorrect because although the induced electromotive force (emf) in both coils is the same, the currents induced in the coils differ due to their varying resistances.
The induced emf, $\varepsilon$, in both coils can be calculated by the formula: $$ \varepsilon = -N \frac{d\phi}{dt} $$ where $\frac{d\phi}{dt}$ is the rate of change of magnetic flux, which is given to be the same for both coils. Therefore, we have: $$ \varepsilon_P = \varepsilon_Q $$
However, the current induced in each coil, denoted $i$, depends on the resistance $R$ of the coil and is given by Ohm's law: $$ i = \frac{\varepsilon}{R} $$ Given the resistivity of iron is greater than that of copper ($\rho_{\text{iron}} > \rho_{\text{copper}}$), the resistance of the iron coil is higher than that of the copper coil: $$ R_{\text{iron}} > R_{\text{copper}} $$
Consequently, the current induced in the iron coil ($Q$) will be less than the current in the copper coil ($P$): $$ i_Q < i_P $$
Therefore, Option D is the incorrect statement as it claims the induced currents are the same, which contradicts the effect of differing resistances on the induced currents in the coils.
A bar magnet has coercivity $4 \times 10^{3} \mathrm{Am}^{-1}$. It is desired to demagnetize it by inserting it inside a solenoid $12 \mathrm{~cm}$ long and having 60 turns. The current that should be sent through the solenoid is
A) $2 \mathrm{~A}$
B) $4 \mathrm{~A}$
C) $6 \mathrm{~A}$
D) $8 \mathrm{~A}$
The correct option is D) $8 \mathrm{~A}$.
We start by noting that the coercivity of the bar magnet is $4 \times 10^{3} \mathrm{Am}^{-1}$. This means the required magnetic intensity, $H$, to demagnetize the magnet is $$ H = 4 \times 10^{3} \mathrm{Am}^{-1}. $$
Given the solenoid has 60 turns and is $12 \mathrm{~cm} = 0.12 \mathrm{~m}$ long, the number of turns per meter, $n$, is calculated as $$ n = \frac{\text{Number of turns}}{\text{Length in meters}} = \frac{60}{0.12} = 500 \text{ turns per meter}. $$
From the formula $H = ni$, where $i$ is the current in amperes through the solenoid, we solve for $i$: $$ i = \frac{H}{n} = \frac{4 \times 10^{3}}{500} = 8.0 \mathrm{~A}. $$
Thus, to demagnetize the bar magnet, a current of 8 A should be sent through the solenoid.
Which of the following is used to locate the direction at a place?
A. Electric Wire
B. Magnetic Compass
C. Capacitor
D. Steel Needle
The correct answer is B. Magnetic Compass.
A magnetic compass is essential for determining direction. When placed on a flat surface, its needle aligns itself along the north-south axis, pointing towards the magnetic north. This feature makes the magnetic compass a crucial tool for travelers, sailors, and navigators who rely on it to determine their course through unfamiliar territories.
Which of the following force is utilized in reducing air pollution by removing dust, soot, and fly-ash particles from the smoke coming out of chimneys of factories?
(a) magnetic force
(b) gravitational force
(c) electrostatic force
(d) frictional force
The correct answer is (c) electrostatic force. This answer describes the force utilized in electrostatic precipitators, which are commonly used to clean industrial emissions. These precipitators work by charging dust and particulate matter in the air using an electrostatic charge. The charged particles are then attracted to and collected on plates or other collection devices with an opposite charge, effectively removing these particles from the emissions before they are released from the chimney. This method significantly reduces air pollution by capturing unwanted particles such as dust, soot, and fly-ash.
State and explain Fleming's right-hand rule.
Fleming's Right-Hand Rule is a useful technique to determine the direction of induced current in a straight conductor. This rule is particularly helpful in situations involving motors and generators where you need to know the orientation of current, magnetic field, and force.
According to Fleming's Right-Hand Rule:
Extend your right hand's thumb, forefinger (index finger), and middle finger such that they are mutually perpendicular to each other.
The forefinger represents the direction of the magnetic field.
The thumb indicates the direction of motion of the conductor, or the direction of the applied force.
The middle finger shows the direction of the induced current.
This configuration helps to visually determine the relationship between the magnetic field, the movement of the conductor, and the resulting direction of the current, based on the orientation of your fingers.
If the number of turns of a circular current-carrying coil is doubled, how will the magnetic field produced by it change?
The magnetic field produced by a current-carrying coil is influenced by several factors, one of which is the number of turns in the coil. The relationship between the number of turns and the magnetic field is directly proportional.
When the number of turns ($N$) is doubled, the magnetic field produced by the coil is also doubled. This effect arises because each turn of the coil contributes equally to the overall magnetic field. Since the current I flowing through each turn is the same and the direction of current in each turn aligns, the additional turns simply enhance the magnetic field generated proportionally. Therefore, if the number of turns is doubled, the resultant magnetic field at any point will be: $$ B_{new} = 2B_{original} $$ where $B_{original}$ is the magnetic field produced by the coil before the number of turns was doubled.
A straight current-carrying conductor is placed in a magnetic field but no force acts on it. What can be the reason behind this?
A. The conductor is parallel to the magnetic field.
B. The conductor is perpendicular to the magnetic field.
C. The conductor is making $60^{\circ}$ with the magnetic field.
D. The conductor is making $45^{\circ}$ with the magnetic field.
The correct answer is A. The conductor is parallel to the magnetic field.
When a straight current-carrying conductor is placed in a magnetic field, the force exerted on the conductor is determined by the orientation of the conductor relative to the direction of the magnetic field. According to Fleming’s left-hand rule, the direction of the force is perpendicular to both the current in the conductor and the magnetic field. However, no force is exerted when the conductor is parallel to the magnetic field, because the angle between the current direction and the magnetic field is zero, resulting in a cross product of zero and hence no force.
An electric bell uses:
A) An electromagnet
B) an electromagnet
C) a compass
D) a piece of lodestone
The correct option is B) an electromagnet.
An electric bell utilizes an electromagnet to intermittently attract a striker, which in turn periodically hits a gong, thus producing sound. This component is pivotal in the functionality of the bell, effectively transforming electrical energy into mechanical movement and sound.
A wire of length l is bent in the form of a circular loop and is suspended in a region of a uniform magnetic field. When a steady current I is passed through the loop, the maximum torque experienced by it is
A) $\frac{l^{2} B}{4 \pi^{2}}$
B) $\frac{B l^{2}}{4 \pi}$
C) $4 \pi^{2} l^{2} B$
D) $\frac{1}{4 \pi B^{2}}$
The correct option is B
$$ \frac{B l^{2}}{4 \pi} $$
The torque experienced by the loop, $ \tau $, is given by: $$ \tau = BIA \sin \theta $$ Where $ \theta $ is the angle between the plane of the loop and the magnetic field. The maximum torque occurs when $ \sin \theta = 1 $, thus: $$ \tau_{\text{max}} = BIA $$
Given the length of the loop is $ l $, the circumference of the circular loop is therefore: $$ l = 2 \pi R \Rightarrow R = \frac{l}{2 \pi} $$
The area of the loop, $ A $, can be calculated using: $$ A = \pi R^2 = \pi \left(\frac{l}{2 \pi}\right)^2 = \frac{l^2}{4 \pi} $$
Substituting $ A $ in the expression for $ \tau_{\text{max}} $, we get: $$ \tau_{\text{max}} = \frac{B l^2 I}{4 \pi} $$
Thus, the correct answer is B: $$ \frac{B l^{2}}{4 \pi} $$
A coil having $N$ turns is wound tightly in the form of a spiral with inner and outer radii 'a' and 'b' respectively. When a current I passes through the coil, the magnetic field at the centre is
A) $\frac{\mu_{0} \mathrm{NI}}{b}$
B) $\frac{2 \mu_{0} \mathrm{NI}}{a}$
C) $\frac{\mu_{0} \mathrm{NI}}{2(b-a)} \ln \frac{b}{a}$
D) $\frac{\mu_{0} \mathrm{N}}{2(b-a)} \ln \frac{b}{a}$
The correct answer is Option C: $\frac{\mu_{0} \mathrm{NI}}{2(b-a)} \ln \frac{b}{a}$
Firstly, let's determine the number of turns per unit width within the coil, which can be given by: $$ \frac{N}{b-a} $$ Now, consider an elemental ring within this coil that has a radius $x$ and an infinitesimal thickness $dx$. The number of turns in this elemental ring can be expressed as: $$ \mathrm{dN} = \frac{N , dx}{b - a} $$
Next, we calculate the magnetic field at the center due to this elemental ring. The expression for the magnetic field ($\mathrm{dB}$) due to a small segment of the coil is: $$ \mathrm{dB} = \frac{\mu_0 (\mathrm{dN}) I}{2x} = \frac{\mu_0 I}{2} \cdot \frac{N , dx}{b - a} \cdot \frac{1}{x} $$
To find the total magnetic field at the center of the coil, we integrate $\mathrm{dB}$ over the entire range of $x$ from $a$ to $b$: $$ \mathrm{B} = \int_{a}^{b} \mathrm{dB} = \int_{a}^{b} \frac{\mu_0 N I}{2(b-a)} \frac{dx}{x} = \frac{\mu_0 N I}{2(b-a)} \ln \frac{b}{a} $$
Thus, the total magnetic field at the center of the spiral coil, when a current $I$ flows through it, is given by: $$ \frac{\mu_0 NI}{2(b-a)} \ln \frac{b}{a} $$ which confirms that Option C is indeed the correct answer.
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