Question

There are two long cylinders of radius $R$ and $2R$ rotating with angular speed of $2\omega$ and $\omega$ (in opposite sense) respectively. If charge density is $-2\sigma$ and $\sigma$ for inner and outer cylinder then magnetic field along the common axis of cylinders is

• A. $6\mu_0 \sigma\omega R$
• B. $-6\mu_0 \sigma\omega R$
• C. $4\mu_0 \sigma\omega R$
• D. $2\mu_0 \sigma\omega R$

Answer 1

Answer: (A)

$6\mu_0 \sigma\omega R$

Answer 2

The magnetic field on the axis of a long rotating charged cylinder is given by $B_z = \mu_0 \sigma' \omega' r$, where $\sigma'$ is the surface charge density, $\omega'$ is the angular velocity, and $r$ is the radius. The direction of $\vec{B}$ is parallel to $\vec{\omega}$ if $\sigma'>0$ and antiparallel if $\sigma'<0$.

Let's assume the common axis is the z-axis. From the figure, the inner cylinder rotates clockwise ($\vec{\omega}_1 = -2\omega \hat{k}$) and the outer cylinder rotates counter-clockwise ($\vec{\omega}_2 = \omega \hat{k}$).

For the inner cylinder: Radius $R$, charge density $\sigma_1 = -2\sigma$. Angular velocity $\omega_1 = -2\omega$. The magnetic field contribution is $B_{z1} = \mu_0 \sigma_1 \omega_1 R = \mu_0 (-2\sigma)(-2\omega)R = 4\mu_0 \sigma\omega R$. Since $\sigma_1$ is negative, the field $\vec{B}_1$ is antiparallel to $\vec{\omega}_1$. As $\vec{\omega}_1$ is in the $-\hat{k}$ direction, $\vec{B}_1$ is in the $+\hat{k}$ direction.

For the outer cylinder: Radius $2R$, charge density $\sigma_2 = \sigma$. Angular velocity $\omega_2 = \omega$. The magnetic field contribution is $B_{z2} = \mu_0 \sigma_2 \omega_2 (2R) = \mu_0 (\sigma)(\omega)(2R) = 2\mu_0 \sigma\omega R$. Since $\sigma_2$ is positive, the field $\vec{B}_2$ is parallel to $\vec{\omega}_2$. As $\vec{\omega}_2$ is in the $+\hat{k}$ direction, $\vec{B}_2$ is in the $+\hat{k}$ direction.

The total magnetic field along the common axis is the sum of the individual fields: $B_{total} = B_{z1} + B_{z2} = 4\mu_0 \sigma\omega R + 2\mu_0 \sigma\omega R = 6\mu_0 \sigma\omega R$.