A particle carrying a charge $q$, moves with a constant angular speed $\omega$ on a circular path. The radius of this path is $r$ & the center is at the origin. How many times does the magnetic field become zero at the point $(2r, 4r)$?

A. 1 time
B. 2 times
C. Never
D. 3 times

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The correct option is B: 2 times.

To determine how often the magnetic field becomes zero at the point $(2r, 4r)$, consider the movement of a particle carrying a charge $q$ with a constant angular speed $\omega$ on a circular path of radius $r$, centered at the origin.

The magnetic field $\vec{B}$ due to a moving charge is given by the formula:

$$ \vec{B} = \frac{\mu_0}{4\pi} \frac{q \vec{v} \times \vec{r}}{r^3} $$


  • $\vec{v}$ is the velocity vector of the particle,
  • $\vec{r}$ is the position vector from the charge to the point where the field is measured,
  • $\mu_0$ is the permeability of free space.

For the magnetic field $\vec{B}$ to become zero, the cross product $\vec{v} \times \vec{r}$ must be zero. This happens when $\vec{v}$ is parallel to $\vec{r}$.

On the circular path, the charge is traveling at a constant angular speed $\omega$, meaning its velocity vector $\vec{v}$ changes direction continuously. However, there are specific moments when $\vec{v}$ aligns with the position vector towards the point of interest $(2r, 4r)$.

In one complete revolution, this alignment happens twice, causing the magnetic field $\vec{B}$ at point $(2r, 4r)$ to be zero at these two specific positions.

Therefore, the magnetic field becomes zero at the point $(2r, 4r)$ 2 times.

Answer: B. 2 times

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