Magnetism and Matter - Class 12 Physics - Chapter 6 - Notes, NCERT Solutions & Extra Questions
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Extra Questions - Magnetism and Matter | NCERT | Physics | Class 12
Material suitable for making a permanent magnet is:
a) Steel
b) Iron
c) Copper
d) Aluminum
The correct answer is option A: Steel.
Materials suitable for making permanent magnets typically exhibit ferromagnetic or ferrimagnetic properties. Common examples of such materials include iron, nickel, cobalt, and certain rare earth metals. Among the options given, steel, which is primarily composed of iron, can retain magnetization and is often used in the manufacturing of permanent magnets. Other materials listed like copper and aluminum are not suitable as they do not exhibit the necessary magnetic properties.
Therefore, steel is the most appropriate choice for making permanent magnets.
At a certain place, the horizontal component of Earth's magnetic field is $\sqrt{3}$ times the vertical component. The angle of dip at that place is
(A) $60^{\circ}$
(B) $45^{\circ}$
(C) $90^{\circ}$
(D) $30^{\circ}$
The question informs us that the horizontal component of the Earth's magnetic field is $\sqrt{3}$ times the vertical component. This is given by the formula:
$$ B_H = \sqrt{3} B_V $$
where $B_H$ is the horizontal component and $B_V$ is the vertical component.
The angle of dip, $\theta$, is defined such that:
$$ \tan \theta = \frac{B_V}{B_H} $$
Substituting the given relation into the equation:
$$ \tan \theta = \frac{B_V}{\sqrt{3} B_V} = \frac{1}{\sqrt{3}} $$
From trigonometric identities, we know that $\tan 30^\circ = \frac{1}{\sqrt{3}}$. Thus, comparing this identity:
$$ \theta = 30^\circ $$
Therefore, the correct answer is (D) $30^\circ$.
If a material is repelled by a magnet, we can infer that the material is a:
A) magnet
B) magnetic material
C) non-magnetic material
The correct answer is C) non-magnetic material.
When a material is repelled by a magnet, it suggests that it does not interact magnetically with the magnet. Generally, materials that interact with magnets either get attracted (if they are ferromagnetic) or repelled (if they are diamagnetic to a very slight extent mostly unnoticeable). Non-magnetic materials, however, typically do not exhibit any magnetic interaction (attraction or repulsion) with magnets. Thus, a material that is visibly repelled by a magnet might suggest it has individual magnet-like properties itself, implying some level of magnetic polarity, but usually, with common observation, non-magnetic materials are those not attracted to magnets. However, exceptionally, if they are repelled, they might possess diamagnetic properties but are generally considered non-magnetic in commonplace terms.
From mining of Magnetite (an ore of iron) $30%$ of the ore gets wasted. Of the remaining ore, only $40%$ is pure iron. If from a mine, the yearly output of iron is 56,000 $\mathrm{kg}$, then what is the quantity of Magnetite (in $\mathrm{kg}$) yearly mined from the mine?
मैग्नेटाइट (लौह-अयस्क) के खनन से $30%$ लौह-अयस्क बर्बाद हो जाता है। शेष अयस्क में केवल $40%$ शुद्ध लोहा है। यदि एक खदान से लोहे का वार्षिक आउटपुट 56,000 $\mathrm{kg}$ है, तो खदान से उत्खनित मैग्नेटाइट की वार्षिक मात्रा (कि.ग्रा. में) क्या है?
A) $2,40,000 \mathrm{~kg}$
B) $1,80,000 \mathrm{~kg}$
C) $2,00,000 \mathrm{~kg}$
D) $2,20,000 \mathrm{~kg}$
The correct answer is C) $200,000 \mathrm{~kg}$.
Let the total quantity of Magnetite ore mined be $x$ kg.
- Ore wasted: $30%$ of $x$, calculated as: $$ \frac{30}{100} \times x = 0.3x $$
- Remaining ore: $70%$ of $x$, calculated as: $$ \frac{70}{100} \times x = 0.7x $$
- Since only $40%$ of the remaining ore is pure iron, the amount of pure iron extracted from the ore can be calculated as: $$ \frac{40}{100} \times 0.7x $$ Simplifying this gives: $$ 0.28x $$
- Given that the yearly output of iron is $56,000$ kg, we set up the following equation and solve for $x$: $$ 0.28x = 56,000 \implies x = \frac{56,000}{0.28} = 200,000 \text{ kg} $$ Hence, the yearly mined quantity of Magnetite from the mine is $200,000$ kg.
The line on the earth's surface joining the points where the earth's magnetic field is horizontal is called:
A) Magnetic meridian
B) Magnetic axis
C) Magnetic line
D) Magnetic equator
The correct answer is D) Magnetic equator.
The Magnetic equator is a specific line on the Earth's surface along which the Earth's magnetic field is perfectly horizontal. At this location, a magnetic needle that is free to move will remain horizontal, pointing neither towards the sky nor the ground, due to the horizontal nature of magnetic lines of force at this line.
Which of the following is/are not matter?
A Love
B Hatred
C Electron
D Air
The correct options are:
- A: Love
- B: Hatred
Explanation:
-
Matter is defined as anything that has mass and occupies space (volume).
-
Love and Hatred are not matter because they are emotions, which do not possess mass or occupy physical space.
-
Electron is considered matter as it has both mass and occupies space.
-
Air is also matter because it consists of gases that have mass and fill the space in which they are contained.
"What is the difference between diamagnetic, paramagnetic, and ferromagnetic?"
Differences Between Diamagnetic, Paramagnetic, and Ferromagnetic Materials
Diamagnetic Material
- A diamagnetic material induces a magnetic response opposite to that of the applied magnetic field, often characterized by a slight repulsion from the magnet.
- These materials, such as silver, copper, and carbon, have permeability values (μr) slightly less than free space (e.g., copper's μr= 0.9999980).
Paramagnetic Material
- Paramagnetic materials align themselves with the magnetic field, showing a weak attraction to magnetic fields.
- Examples include aluminium and air, which have permeability slightly greater than free space (e.g., air's μr = 1.0000004).
Ferromagnetic Material
- Ferromagnetic materials, a subset of paramagnetic materials, are strongly attracted to magnetic fields and can retain magnetic properties even after the external field is removed.
- Common examples include iron, steel, cobalt, and their alloys, with relative permeability values much higher than free space, typically extending into the hundreds or thousands.
The impacts of both diamagnetic and paramagnetic properties are generally considered very small, so materials with these characteristics are frequently described as non-magnetic.
Comparative Properties Table
PROPERTIES | FERROMAGNETIC MATERIALS | PARAMAGNETIC MATERIALS | DIAMAGNETIC MATERIALS |
---|---|---|---|
State | Solid | Can be solid, liquid, or gas | Can be solid, liquid, or gas |
Effect of Magnet | Strongly attracted | Weakly attracted | Weakly repelled |
Behavior under non-uniform field | Tend to move from low to high field region | Tend to move from low to high field region | Tend to move from high to low region |
Behavior under external field | Preserve magnetic properties post-field | Do not preserve magnetic properties | Do not preserve magnetic properties |
Effect of Temperature | Becomes paramagnetic above the Curie point | Becomes diamagnetic with temperature rise | No effect |
Permeability | Very high | Slightly greater than unity | Slightly less than unity |
Susceptibility | Very high and positive | Slightly greater than unity and positive | Slightly less than unity and negative |
Examples | Iron, Nickel, Cobalt | Lithium, Tantalum, Magnesium | Copper, Silver, Gold |
This table and descriptions convey the fundamental behaviors and characteristics, highlighting the differences between diamagnetic, paramagnetic, and ferromagnetic materials across various properties.
A body of mass $\sqrt{3} \mathrm{~kg}$ is suspended by a string to a rigid support. The body is pulled horizontally by a force $F$ until the string makes an angle of $30^{\circ}$ with the vertical. The value of $F$ and tension in the string are.
A 9.8 $\mathrm{N}$, 9.8 $\mathrm{N}$
B. 9.8 N, 19.6 N
C. 19.6 N, 19.6 N
D. 19.6 N, 9.8 N
To solve this problem, we need to analyze the forces acting on the body of mass $\sqrt{3} \text{ kg}$ suspended by a string when it is pulled horizontally by a force $F$ until the string makes an angle of $30^{\circ}$ with the vertical.
Step-by-Step :
Identify the Forces:
The weight of the body, $mg$, acts downwards.
The tension in the string, $T$, acts along the string.
The horizontal force, $F$, pulls the body to the side.
Resolve the Tension:
The tension $T$ can be resolved into two components:
Vertical component: $T \cos \theta$
Horizontal component: $T \sin \theta$
For $\theta = 30^{\circ}$:
$T \cos 30^{\circ} = T \frac{\sqrt{3}}{2}$
$T \sin 30^{\circ} = T \frac{1}{2}$
Equilibrium Conditions:
Vertical equilibrium:$$ T \cos 30^{\circ} = mg \rightarrow T \frac{\sqrt{3}}{2} = \sqrt{3} \times 9.8 \text{ m/s}^2 $$
Solving for $T$: $$ T \frac{\sqrt{3}}{2} = \sqrt{3} \times 9.8 \implies T = \frac{2 \times \sqrt{3} \times 9.8}{\sqrt{3}} = 2 \times 9.8 = 19.6 \text{ N} $$
Horizontal equilibrium:$$ T \sin 30^{\circ} = F \rightarrow T \frac{1}{2} = F \implies F = \frac{T}{2} = \frac{19.6}{2} = 9.8 \text{ N} $$
Final :
The value of the horizontal force $F$ is 9.8 N.
The tension in the string $T$ is 19.6 N.
Thus, the correct option is:
B) 9.8 N, 19.6 N
In the CGS system, the magnitude of the force is 100 dynes. In another system where the fundamental physical quantities are kilogram, meter, and minute, find the magnitude of the force.
A. 0.036
B. 0.36
C. 3.6
D. 36
To find the magnitude of the force 100 dynes in another system where the fundamental physical quantities are kilogram, meter, and minute, we proceed as follows:
Step-by-Step
Understanding the Given Units:
1 dyne in the CGS system is equal to $1 , \text{g} \cdot \text{cm}/\text{s}^2$.
Conversion Factors:
$1 , \text{g} = 10^{-3} , \text{kg}$
$1 , \text{cm} = 10^{-2} , \text{m}$
$1 , \text{s} = \frac{1}{60} , \text{minute}$
Convert 1 dyne to the New System:
First, rewrite 1 dyne with the new units: $$1, \text{dyne} = 10^{-3} , \text{kg} \cdot 10^{-2} , \text{m} \cdot \left(\frac{60}{1 , \text{minute}}\right)^2$$
Calculate the Magnitude:
Simplify the expression: $$1, \text{dyne} = 10^{-3} , \text{kg} \cdot 10^{-2} , \text{m} \cdot 3600 , \text{minute}^{-2}$$ $$1 , \text{dyne} = 10^{-3} \times 10^{-2} \times 3600 , \text{kg} \cdot \text{m}/\text{minute}^2$$ $$1 , \text{dyne} = 3.6 \times 10^{-5} , \text{kg} \cdot \text{m}/\text{minute}^2$$
Convert 100 Dynes:
Multiply the result by 100: $$100 , \text{dynes} = 100 \times 3.6 \times 10^{-5} , \text{kg} \cdot \text{m}/\text{minute}^2$$ $$100 , \text{dynes} = 3.6 \times 10^{-3} , \text{kg} \cdot \text{m}/\text{minute}^2$$
Thus, the magnitude of the force in the new system is 3.6 kg·m/minute².
So, the correct answer is C. 3.6.
In a simple pendulum experiment, length is measured as 31.4 cm with an accuracy of 1 mm. The time for 100 oscillations of the pendulum is 112.0 s with an accuracy of 0.1 s. The percentage accuracy in g is 1.3%.
In this problem, we need to find the percentage accuracy in ( g ) (acceleration due to gravity) in a simple pendulum experiment. Let's break down the given data and the steps to solve this:
Given Data:
Length (( l )): ( 31.4 ) cm with an accuracy of ( 1 ) mm.
Time for 100 oscillations (( t_{100} )): ( 112.0 ) s with an accuracy of ( 0.1 ) s.
Step-by-Step :
Convert Length and Accuracy to Consistent Units:
Length ( l = 31.4 ) cm (convert to meters: ( l = 0.314 ) m)
Least count for length ( \Delta l = 0.1 ) cm (convert to meters: ( \Delta l = 0.001 ) m)
Calculate the Time Period:
Time period ( T ) for one oscillation: [ T = \frac{t_{100}}{100} = \frac{112.0}{100} = 1.12 \text{ s} ]
Accuracy in time period ( \Delta T = \frac{0.1}{100} = 0.001 \text{ s} )
Percentage Uncertainties in Length and Time Period:
Percentage accuracy in length ( \frac{\Delta l}{l} \times 100 % ): [ \frac{0.001}{0.314} \times 100 \approx 0.32% ]
Percentage accuracy in time period ( \frac{\Delta T}{T} \times 100 % ): [ \frac{0.001}{1.12} \times 100 \approx 0.89% ]
Calculate the Percentage Accuracy in ( g ):According to the formula for the time period of a simple pendulum: [ T = 2\pi \sqrt{\frac{l}{g}} ] Solving for ( g ), we get: [ g = \frac{4\pi^2 l}{T^2} ] The percentage uncertainty in ( g ) is given by: [ \frac{\Delta g}{g} \times 100 = \left( \frac{\Delta l}{l} \times 100 \right) + 2 \left( \frac{\Delta T}{T} \times 100 \right) ] Substituting the values, we get: [ \frac{\Delta g}{g} \times 100 = 0.32% + 2 \times 0.89% \approx 2.10% ]
Conclusion:
The percentage accuracy in ( g ) is ${2.1%}$.
A rectangular metal slab of mass 33.333 g has a length of 8.0 cm, a breadth of 5.0 cm, and a thickness of 1 mm. The mass is measured with accuracy up to 1 mg with a sensitive balance. The length and breadth are measured with a vernier calipers having a least count of 0.01 cm. The thickness is measured with a screw gauge of least count 0.01 mm.
To calculate the percentage accuracy in density of the metal slab given the measurements provided, we follow these steps:
Given Data:
Mass $m = 33.333 , \text{g}$
Length $l = 8.0 , \text{cm}$
Breadth $b = 5.0 , \text{cm}$
Thickness $t = 1 , \text{mm} = 0.1 , \text{cm}$
Measurement Accuracies:
Mass accuracy: $\Delta m = 0.001 , \text{g}$ (1 mg)
Length accuracy: $\Delta l = 0.01 , \text{cm}$
Breadth accuracy: $\Delta b = 0.01 , \text{cm}$
Thickness accuracy: $\Delta t = 0.01 , \text{mm} = 0.001 , \text{cm}$
Formulae Used:
Density ($\rho$): $$ \rho = \frac{m}{V} = \frac{m}{l \times b \times t} $$
Volume ($V$): $$ V = l \times b \times t $$
Error in Density: $$ \frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + \frac{\Delta l}{l} + \frac{\Delta b}{b} + \frac{\Delta t}{t} $$
Calculating Initial Density:
Volume: $$ V = l \times b \times t = 8.0 , \text{cm} \times 5.0 , \text{cm} \times 0.1 , \text{cm} = 4.0 , \text{cm}^3 $$
Density: $$ \rho = \frac{m}{V} = \frac{33.333 , \text{g}}{4.0 , \text{cm}^3} = 8.33325 , \text{g/cm}^3 $$
Calculating Error in Density:
Relative errors: $$ \frac{\Delta m}{m} = \frac{0.001}{33.333} \approx 3.0003 \times 10^{-5} $$
$$ \frac{\Delta l}{l} = \frac{0.01}{8.0} = 0.00125 $$
$$ \frac{\Delta b}{b} = \frac{0.01}{5.0} = 0.002 $$
$$ \frac{\Delta t}{t} = \frac{0.001}{0.1} = 0.01 $$
Summing relative errors: $$ \frac{\Delta \rho}{\rho} = 3.0003 \times 10^{-5} + 0.00125 + 0.002 + 0.01 = 0.013 $$
Percentage Error in Density: $$ \text{Percentage Error} = \left( \frac{\Delta \rho}{\rho} \right) \times 100 % = 0.013 \times 100 % = 1.3 % $$
Final Answer:
Therefore, the percentage accuracy in density from the given measurements is 1.3%.
A body of weight $20 \text{ N}$ is on a horizontal surface. The minimum force applied to pull it when the applied force makes an angle $60^\circ$ with the horizontal (where the angle of friction $=30^\circ$) is:
A $20 \text{ N}$.
B. $20 \sqrt{3} \text{N}$
C. $\frac{20}{\sqrt{3}} ~\text{N}$
To find the minimum force required to pull a body of weight $20 , \text{N}$ on a horizontal surface when the applied force makes an angle of $60^\circ$ with the horizontal, we need to consider the forces acting on the body and resolve them into components.
Given:
Weight of the body, $W = 20 , \text{N}$
Angle of applied force with the horizontal, $\theta = 60^\circ$
Angle of friction, $\phi = 30^\circ$
Steps:
Coefficient of Friction:We can find the coefficient of friction ($\mu$) using the angle of friction. $$ \tan(\phi) = \mu $$ Given $\phi = 30^\circ$: $$\tan(30^\circ) = \frac{1}{\sqrt{3}} ] Therefore, [ \mu = \frac{1}{\sqrt{3}} $$
Force Resolution:For pulling the body, the minimum force ($F$) required can be found by resolving the applied force into its horizontal and vertical components: $$ F_x = F \cos(\theta) $$ $$ F_y = F \sin(\theta) $$
Normal Force Adjustment:The normal force ($N$) is adjusted by the vertical component of the pulling force: $$ N = W - F \sin(\theta) $$
The minimum force required to overcome friction will balance the horizontal component of the pulling force with the frictional force: $$ F \cos(\theta) = \mu N $$ Substituting $N$: $$ F \cos(60^\circ) = \frac{1}{\sqrt{3}} (W - F \sin(60^\circ)) $$ Simplify the trigonometric values: $$ F \cdot 0.5 = \frac{1}{\sqrt{3}} (20 - F \cdot \frac{\sqrt{3}}{2}) $$ $$ F \cdot 0.5 = \frac{1}{\sqrt{3}} (20 - 0.866F) $$ $$ F \cdot 0.5 = \frac{20}{\sqrt{3}} - \frac{0.866F}{\sqrt{3}} $$
Solving for $F$:Combine terms and solve for $F$: $$ 0.5F + \frac{0.866}{\sqrt{3}}F = \frac{20}{\sqrt{3}} $$ Simplify: $$ F \left(0.5 + \frac{0.866}{\sqrt{3}}\right) = \frac{20}{\sqrt{3}} $$ Calculate the coefficient: $$ F \left(0.5 + 0.5\right) = \frac{20}{\sqrt{3}} $$ $$ F = \frac{20}{\sqrt{3}} $$
Therefore, the minimum force required to pull the body is:
$$ \boxed{\frac{20}{\sqrt{3}} , \text{N}} $$
A block of mass 5 kg is lying on a rough horizontal surface. The coefficient of static and kinetic friction are 0.3 and 0.1 and g = 10 m/s². The frictional force on the block is
25 N
5 N
10 N
Zero
To determine the frictional force on a block of mass 5 kg lying on a rough horizontal surface, given the coefficients of static and kinetic friction and the acceleration due to gravity, let's go through the details step-by-step.
Given:
Mass of the block,$m = 5 , \text{kg} $
Coefficient of static friction, $ \mu_s = 0.3 $
Coefficient of kinetic friction, $\mu_k = 0.1$
Gravitational acceleration, $g = 10 , \text{m/s}^2 $
Explanation:
Normal Force: On a horizontal surface, the normal force $ N $ is equal to the gravitational force acting on the block: $$ N = mg = 5 , \text{kg} \times 10 , \text{m/s}^2 = 50 , \text{N} $$
Static and Kinetic Friction:
Maximum static friction force: $$ f_s^{\text{max}} = \mu_s N = 0.3 \times 50 , \text{N} = 15 , \text{N} $$
Kinetic friction force: $$ f_k = \mu_k N = 0.1 \times 50 , \text{N} = 5 , \text{N} $$
Determining the Frictional Force: The friction force depends on whether there is an external force applied to the block. Since the problem does not mention any external force:
No external force means there is no motion; hence, no frictional force is needed to counteract any movement.
Conclusion:
Since no external force is applied to the block, resulting in no need to oppose any force, the frictional force on the block is:
$$ \text{Frictional Force} = 0 , \text{N} $$
Hence, the correct option is zero.
Final Answer: Zero
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A short bar magnet placed with its axis at $30^{\circ}$ with a uniform external magnetic field of $0.25 \mathrm{~T}$ experiences a torque of magnitude equal to $4.5 \times 10^{-2} \mathrm{~J}$. What is the magnitude of magnetic moment of the magnet?
The magnitude of the magnetic moment $ \mathbf{m} $ of the magnet can be calculated using the formula for the torque experienced by a magnetic dipole in a uniform magnetic field:
$$ \tau = m B \sin \theta $$
Given:
Torque, ( \tau = 4.5 \times 10^{-2} , \text{J} )
Magnetic field, ( B = 0.25 , \text{T} )
Angle, ( \theta = 30^{\circ} )
Rearranging the formula, we get:
$$ m = \frac{\tau}{B \sin \theta} $$
Substitute the given values:
$$ m = \frac{4.5 \times 10^{-2}}{0.25 \times \sin 30^{\circ}} $$
And solving this, we find:
$$ m = 0.36 , \text{A} \cdot \text{m}^2 $$
Thus, the magnitude of the magnetic moment of the magnet is 0.36 A·m(^2).
A short bar magnet of magnetic moment $\mathrm{m}=0.32 \mathrm{JT}^{-1}$ is placed in a uniform magnetic field of $0.15 \mathrm{~T}$. If the bar is free to rotate in the plane of the field, which orientation would correspond to its (a) stable, and (b) unstable equilibrium? What is the potential energy of the magnet in each case?
Results
Stable Equilibrium (( \theta = 0^\circ )): [ U_{\text{stable}} = - (0.32 , \mathrm{J/T}) \cdot (0.15 , \mathrm{T}) \cdot \cos(0^\circ) = -0.048 , \mathrm{J} ]
Unstable Equilibrium (( \theta = 180^\circ )): [ U_{\text{unstable}} = - (0.32 , \mathrm{J/T}) \cdot (0.15 , \mathrm{T}) \cdot \cos(180^\circ) = 0.048 , \mathrm{J} ]
Summary
Stable Equilibrium: The bar magnet aligns with the magnetic field (( \theta = 0^\circ )) and the potential energy is -0.048 J.
Unstable Equilibrium: The bar magnet is anti-aligned with the magnetic field (( \theta = 180^\circ )) and the potential energy is 0.048 J.
A closely wound solenoid of 800 turns and area of cross section $2.5 \times 10^{-4} \mathrm{~m}^{2}$ carries a current of 3.0 A. Explain the sense in which the solenoid acts like a bar magnet. What is its associated magnetic moment?
The magnetic moment $m $ of the solenoid is:
$$ m = 0.6 , \mathrm{A \cdot m^2} $$
Thus, the solenoid behaves like a bar magnet with an associated magnetic moment of 0.6 A·m².
If the solenoid in Exercise 5.5 is free to turn about the vertical direction and a uniform horizontal magnetic field of $0.25 \mathrm{~T}$ is applied, what is the magnitude of torque on the solenoid when its axis makes an angle of $30^{\circ}$ with the direction of applied field?
The magnitude of the torque on the solenoid is:
[ \tau = 785.4 , \text{N m} ]
Thus, the magnitude of the torque on the solenoid when its axis makes an angle of (30^\circ) with the direction of the applied field is 785.4 N m.
A bar magnet of magnetic moment $1.5 \mathrm{~J} \mathrm{~T}^{-1}$ lies aligned with the direction of a uniform magnetic field of $0.22 \mathrm{~T}$.
(a) What is the amount of work required by an external torque to turn the magnet so as to align its magnetic moment: (i) normal to the field direction, (ii) opposite to the field direction?
(b) What is the torque on the magnet in cases (i) and (ii)?
(a) Work required by an external torque
To turn the magnet to align its magnetic moment normal to the field direction:
Work required, ( \Delta U_m = mB ): [ 1.5 \mathrm{~J} \mathrm{~T}^{-1} \times 0.22 \mathrm{~T} = 0.33 \mathrm{~J} ]
To turn the magnet to align its magnetic moment opposite to the field direction:
Work required, ( \Delta U_m = 2mB ): [ 2 \times 1.5 \mathrm{~J} \mathrm{~T}^{-1} \times 0.22 \mathrm{~T} = 0.66 \mathrm{~J} ]
(b) Torque on the magnet
Normal to the field direction ($ \theta = 90^\circ $):
Torque, ( \tau = mB \sin 90^\circ = mB ): [ 1.5 \mathrm{~J} \mathrm{~T}^{-1} \times 0.22 \mathrm{~T} = 0.33 \mathrm{~N} \mathrm{~m} ]
Opposite to the field direction ($ \theta = 180^\circ $):
Torque, $ \tau = mB \sin 180^\circ = 0 \mathrm{~N} \mathrm{~m} $:
Summary
(a) Work Required
Normal to the field: $ \boldsymbol{0.33 \text{ J}} $
Opposite to the field: $\boldsymbol{0.66 \text{ J}} $
(b) Torque on the Magnet
Normal to the field: $\boldsymbol{0.33 \text{ N} \cdot \text{m}}$
Opposite to the field: $\boldsymbol{0 \text{ N} \cdot \text{m}} $
A closely wound solenoid of 2000 turns and area of cross-section $1.6 \times 10^{-4} \mathrm{~m}^{2}$, carrying a current of $4.0 \mathrm{~A}$, is suspended through its centre allowing it to turn in a horizontal plane.
(a) What is the magnetic moment associated with the solenoid?
(b) What is the force and torque on the solenoid if a uniform horizontal magnetic field of $7.5 \times 10^{-2} \mathrm{~T}$ is set up at an angle of $30^{\circ}$ with the axis of the solenoid?
(a) Magnetic Moment
The magnetic moment ( m ) of a solenoid is given by: $$ m = n I A $$ where
( n ) is the number of turns,
( I ) is the current,
( A ) is the area of cross-section.
Given:
$ n = 2000 $,
$ I = 4.0 , \mathrm{A}$,
$ A = 1.6 \times 10^{-4} , \mathrm{m}^2 $.
Hence, $$ m = 2000 \times 4.0 \times 1.6 \times 10^{-4} , \mathrm{A \cdot m^2} $$ $$ m = 1.28 , \mathrm{A \cdot m^2} $$
So, the magnetic moment associated with the solenoid is: $$ \mathbf{m} = 1.28 , \mathrm{A \cdot m^2} $$
(b) Force and Torque on the Solenoid
Force
In a uniform magnetic field, the net force on a solenoid is zero because the magnetic forces on opposite sides cancel out each other. Therefore, $$ \text{Net Force} = 0 $$
Torque
The torque $ \tau$ on a solenoid in a magnetic field is given by: $$ \tau = m B \sin \theta $$ where
( m ) is the magnetic moment,
( B ) is the magnetic field,
( \theta ) is the angle between the magnetic moment and the magnetic field direction.
Given:
( m = 1.28 , \mathrm{A \cdot m^2} ),
( B = 7.5 \times 10^{-2} , \mathrm{T} ),
( \theta = 30^{\circ} ).
Let's calculate this:
$$ \tau = 1.28 \times 7.5 \times 10^{-2} \times \sin 30^\circ $$
Using ( \sin 30^\circ = 0.5 ),
$$ \tau = 1.28 \times 7.5 \times 10^{-2} \times 0.5 $$
$1.28 7.5 10^-2 * 0.5$
Now we sum up the results:
[ \tau = 1.28 \times 7.5 \times 10^{-2} \times 0.5 = 0.048 , \mathrm{N \cdot m} ]
So, the torque on the solenoid is: $$ \mathbf{\tau}= 0.048 , \mathrm{N \cdot m} $$
To summarize:
Magnetic Moment: $ 1.28 , \mathrm{A \cdot m^2} $
Force: $ 0 , \mathrm{N} $ (no net force in a uniform magnetic field)
Torque:$ 0.048 , \mathrm{N \cdot m} $
A short bar magnet has a magnetic moment of $0.48 \mathrm{~J} \mathrm{~T}^{-1}$. Give the direction and magnitude of the magnetic field produced by the magnet at a distance of $10 \mathrm{~cm}$ from the centre of the magnet on (a) the axis, (b) the equatorial lines (normal bisector) of the magnet.
Here are the calculations for the magnetic field produced by the short bar magnet:
(a) Magnetic field on the axis
The magnetic field at a distance of (0.1 , \mathrm{m}) from the center along the axis of the magnet is: $$ B_{\text{axis}} = \frac{\mu_0}{4 \pi} \frac{2m}{r^3} $$ From the calculation, we have: $$ B_{\text{axis}} = 0.00121 , \mathrm{T} \quad (\text{or } 1.21 , \mathrm{mT}) $$
(b) Magnetic field on the equatorial lines (normal bisector)
The magnetic field at a distance of (0.1 , \mathrm{m}) from the center along the equatorial line is: $$ B_{\text{equator}} = \frac{\mu_0}{4 \pi} \frac{m}{r^3} $$ From the calculation, we have: $$ B_{\text{equator}} = 0.0006 , \mathrm{T} \quad (\text{or } 0.6 , \mathrm{mT}) $$
Directions
On the axis: The magnetic field at this point will be directed along the axis of the magnet. If you are at the north pole of the magnet, this field will point away from the magnet, and if you are at the south pole, it will point towards the magnet.
On the equatorial lines: The magnetic field will be perpendicular to the axis of the magnet and will point from the north to the south pole, forming closed loops around the magnet.
Summary
Magnitude of the magnetic field on the axis: $1.21 , \mathrm{mT}$
Magnitude of the magnetic field on the equatorial line: $0.6 , \mathrm{mT}$
The field along the axis is stronger than the field along the equatorial line at the same distance from the center of the magnet.
A short bar magnet placed in a horizontal plane has its axis aligned along the magnetic north-south direction. Null points are found on the axis of the magnet at $14 \mathrm{~cm}$ from the centre of the magnet. The earth's magnetic field at the place is $0.36 \mathrm{G}$ and the angle of dip is zero. What is the total magnetic field on the normal bisector of the magnet at the same distance as the null-point (i.e., $14 \mathrm{~cm}$ ) from the centre of the magnet? (At null points, field due to a magnet is equal and opposite to the horizontal component of earth's magnetic field.)
The total magnetic field on the normal bisector of the magnet at a distance of $ 0.14\ \mathrm{m} $ from the center of the magnet is approximately:
[ B_E \approx 0.000018\ \mathrm{T} = 1.8 \times 10^{-5}\ \mathrm{T} ]
So, the total magnetic field on the normal bisector at the same distance as the null-point is $ 1.8 \times 10^{-5}\ \mathrm{T} $.
If the bar magnet in exercise 5.13 is turned around by $180^{\circ}$, where will the new null points be located?
Let's break down the equation step by step and then solve for $r$:
Original Magnetic Field Equation
For a bar magnet, the axial magnetic field at a distance $r$ from its center is given by: [ B_m = \frac{\mu_0}{4 \pi} \frac{2 M}{r^3} ]
Null Point Condition
At the null point, the magnetic field due to the bar magnet ($B_m$) is equal and opposite to the Earth's horizontal magnetic field ($B_H$): [ \frac{\mu_0}{4 \pi} \frac{2 M}{r^3} = B_H ]
Solving for ( r )
Re-arranging the equation to solve for ( r ): [ r^3 = \frac{\mu_0}{4 \pi} \frac{2 M}{B_H} ] [ r = \left( \frac{\mu_0}{4 \pi} \frac{2 M}{B_H} \right)^{1/3} ]
We will now use this relation to solve for ( r ) using the given constants:
$\mu_0 $ (permittivity of vacuum) = $4\pi \times 10^{-7} , \text{N/A}^2$
$ M $ (magnetic moment of the bar magnet) - requires value
$ B_H $ (horizontal component of Earth's magnetic field) - requires value
Assuming no specific values are given, a general setup would be to compute the distance without substituting those specific values.
Finding New Null Points
When the bar magnet is turned around by $180^\circ$, the null points shift along the equatorial line but remain at the same distance calculated above.
Let’s consider the scenario in general context for any values for $M$ and $B_H$. If specific values are needed, please provide them and we can compute numerical results.
In general, if the bar magnet is originally oriented such that:
Null points are at distance ( r ) on either side along the equatorial line (Left null point: $-r$ from center, Right null point: $r$ from center)
When flipped by $180^\circ$:
The positions of the null points will also be at $r$ on either side of the magnet but reversed: (New Left null point: $r$, New Right null point: $-r$)
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Notes - Magnetism and Matter | Class 12 NCERT | Physics
Magnetism and Matter Class 12 Notes: Comprehensive Study Guide
Introduction
Understanding magnetism and how matter interacts with magnetic fields is crucial for both academic studies and practical applications. This guide covers essential concepts in magnetism for Class 12 students, including the nature of magnetic fields, the behaviour of bar magnets, and types of magnetic materials.
Understanding Magnetism and Matter
Definition of Magnetism
Magnetism refers to the force exerted by magnets when they attract or repel each other. It is caused by the motion of electric charges. In essence, a magnetic field represents a region where a magnetic force can be detected.
Importance of Magnetic Fields in Nature
Magnetic fields are pervasive in nature. From the expansive magnetic fields of galaxies to the tiny magnetic fields around atoms, magnetism influences various phenomena. The Earth's magnetism, for instance, has existed long before human evolution and plays a vital role in navigation and animal behaviour.
Historical Background and Evolution of Magnetism Studies
The term "magnet" originates from Magnesia, a region in Greece known for its natural magnetic rocks discovered as early as 600 BC. The scientific study of magnetism advanced significantly with contributions from scientists like Oersted, Ampere, Biot, and Savart in the 19th century.
Earth's Magnetism
Basics of Earth's Magnetic Field
The Earth itself acts as a giant magnet with a magnetic field resembling that of a bar magnet. This field roughly aligns with the planet's geographic poles, running from the geographic south pole to the geographic north pole.
Directional Properties of Bar Magnets
When bar magnets are freely suspended, they align themselves in the north-south direction. The north-seeking end is termed the north pole, and the south-seeking end is termed the south pole. This directional alignment is due to Earth's magnetic field influence.
Fundamental Properties of Magnets
North and South Poles of Magnets
Magnets have two poles: north and south. Identical poles repel each other, while opposite poles attract. It is impossible to isolate a single magnetic pole (north or south) since cutting a bar magnet results in two smaller magnets, each having both poles.
Magnetic Monopoles and Their Existence
Contrary to electrons or protons, which can exist independently with positive or negative charges, magnetic monopoles (isolated north or south poles) do not exist. This phenomenon stems from the intrinsic nature of magnetic fields.
Magnetisation of Iron and its Alloys
Iron and its alloys can be magnetised by aligning their internal magnetic domains using external magnetic fields. This process induces a net magnetic moment within the material, converting it into a magnet.
Magnetic Field Lines
Visualising Magnetic Field Lines
Magnetic field lines provide a visual representation of magnetic fields. They form continuous loops from the north to the south pole outside the magnet. The density of these lines indicates the field's strength.
Bar magnet with magnetic field lines flowing from the north pole to the south pole.
Behavioural Patterns of Bar Magnets in External Fields
When a bar magnet is placed in an external magnetic field, it experiences a torque aligning it with the field. The torque is calculated as (\tau = \mathbf{m} \times \mathbf{B}), where (\mathbf{m}) is the magnetic moment and (\mathbf{B}) is the magnetic field.
Gauss’s Law of Magnetism
Explanation of Gauss’s Law
Gauss's law for magnetism states that the net magnetic flux through any closed surface is zero. This indicates that magnetic field lines neither begin nor end but instead form continuous loops.
Applications and Implications
This law implies that magnetic monopoles do not exist and helps in understanding the nature of magnetic fields in different configurations.
Types of Magnetic Materials
Description of Paramagnetism
Paramagnetic materials exhibit weak magnetisation in the presence of an external magnetic field, aligning parallel to the field. They are attracted to regions of stronger magnetic fields.
Description of Diamagnetism
Diamagnetic materials develop a magnetic field in opposition to an externally applied magnetic field, causing repulsion. This phenomenon is due to changes in orbital motion of electrons induced by the external field.
Description of Ferromagnetism
Ferromagnetic materials show strong magnetisation in the direction of an external magnetic field. They have domains—regions with uniformly aligned magnetic moments. Under an external field, these domains align themselves to enhance the overall magnetic field.
Electromagnets vs. Permanent Magnets
Creating Electromagnets
Electromagnets are created by circulating electric current through a coil of wire, often wrapped around a ferromagnetic core, which magnifies the produced magnetic field.
Properties of Permanent Magnets
Permanent magnets retain their magnetisation after the external magnetic field is removed. Materials like steel and certain alloys are often used to make permanent magnets.
Magnetic Moment and Calculations
Equating Bar Magnets to Solenoids
A bar magnet can be considered akin to a solenoid with circulating currents producing a similar magnetic field. Cutting a solenoid or a bar magnet results in smaller magnets, each with both north and south poles.
Calculation of Axial Fields of Bar Magnets
The magnetic field at a distance (r) from a bar magnet is expressed as: [ B = \frac{\mu_{0}}{4 \pi} \frac{2m}{r^3} ] where (m) is the magnetic moment of the bar magnet.
Special Magnetic Properties
Behaviour of Magnetic Materials at Different Conditions
Different materials react uniquely to magnetic fields. For instance, superconductors exhibit perfect diamagnetism by expelling all magnetic field lines, a phenomenon known as the Meissner effect.
Example Calculations with Bar Magnets and Solenoids
Consider a solenoid of 1000 turns per metre carrying a 2 A current with a core of relative permeability 400. The magnetic field can be calculated as: [ B = \mu_{0} \mu_{r} n I = (4 \pi \times 10^{-7})(400)(1000)(2) = 1 T ]
Advanced Concepts in Magnetism
Dipole in a Uniform Magnetic Field
A magnetic dipole in a uniform magnetic field experiences torque aligning it with the field. The potential energy of this system is given by ( U = -\mathbf{m} \cdot \mathbf{B} ).
Magnetic Potential Energy
The magnetic potential energy equation ( U_m = -m B \cos \theta ) shows that the system is most stable when the dipole is aligned with the field.
Electromagnetic Analogies
Comparing electric and magnetic fields: [ \mathbf{E} \rightarrow \mathbf{B}, \mathbf{p} \rightarrow \mathbf{m}, \frac{1}{4 \pi \epsilon_0} \rightarrow \frac{\mu_0}{4 \pi} ]
Practical Applications and Examples
Cutting Bar Magnets: Effects and Outcomes
When a bar magnet is cut transversely or along its length, each piece forms a new magnet with its own north and south poles.
Forces and Torques in Magnetic Fields
Magnetised objects in a magnetic field experience forces and torques when interacting with the field. An iron nail near a magnet experiences both due to induced magnetisation.
Identifying Magnetised Materials
To identify whether an iron bar is magnetised, bring its ends near another bar. If repulsion is observed, both are magnetised; if always attracted, only one is magnetised.
Conclusion
Magnetism and the behaviour of magnetic materials are fundamental topics in physics. Understanding these concepts is not only essential for academic success in Class 12 but also provides a foundation for further studies in science and engineering.
Additional Resources
Further Reading and Problem Sets
- Textbooks and problem collections for deeper insights and practice.
Video Lectures and Tutorials
- Online resources and video lessons to supplement reading material.
flowchart LR
A[Magnetism Basics] --> B[Earth's Magnetism]
B --> C[N-S Alignment]
A --> D[Magnetic Field Lines]
D --> E[Bar Magnet Behaviour]
A --> F[Types of Materials]
F --> G[Paramagnetic]
F --> H[Diamagnetic]
F --> I[Ferromagnetic]
A --> J[Gauss's Law]
A --> K[Electromagnets]
A --> L[Permanent Magnets]
A --> M[Applications]
M --> N[Cutting Magnets]
M --> O[Forces/Torques]
M --> P[Identification]
This article delves into the principles of magnetism and its interaction with matter, providing a comprehensive review suitable for Class 12 students preparing for their exams.
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