Question

A perfectly conducting spherical shell of radius rotates about the z axis with angular velocity $\mathbf{\omega }$, in a uniform magnetic field $\mathbf{B}\mathbf{=}{\mathbf{B}}_{\mathbf{0}}\hat{\mathbf{Z}}$. Calculate the emf developed between the “north pole” and the equator. Answer:localid="1658295408106" $\left[\frac{1}{2}{B}_0{\mathrm{\omega \alpha }}^2\right]$.

Short answer

The emf developed is$\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{B}}_{\mathbf{0}}{\mathbf{\omega \alpha }}^{\mathbf{2}}$.

Step-by-step solution

Given information

The radius ofspherical shellis, a .

The spherical shell rotates about the z axis.

The angular velocity of rotation is, $\omega$.

The uniform magnetic field is, $\mathrm{B}={\mathrm{B}}_0\hat{\mathrm{z}}$.

Magnetic force

As a unit charge moves through a magnetic field then it experiences a certain amount of force. The force experience by the unit charge is described as the ‘magnetic force’.

The magnetic force on a unit charge is equal to the cross product between the velocity of charge and the magnetic field vectors.

Determine the emf developed

The linear velocity of the unit charge on the spherical shell is,

$\mathrm{v}=\mathrm{\omega \alpha sin}\mathrm{\theta }\hat{\mathrm{\varphi }}$.

The formula for the force (f) exerted by magnetic field (B) on a unit charge moving with velocity (v) is given by,

$$\begin{gathered} \mathrm{f}=\mathrm{v}\times \mathrm{B} \\ \mathrm{f}=\mathrm{\omega \alpha }\sin \mathrm{\theta }\hat{\mathrm{\varphi }}\times {\mathrm{B}}_0\hat{\mathrm{z}} \\ \mathrm{f}={\mathrm{\omega \alpha B}}_0\sin \mathrm{\theta }\left(\hat{\mathrm{\varphi }}\times \hat{\mathrm{z}}\right) \end{gathered}$$

Then the formula for the emf developed between the “north pole”$\theta =0$and the equator $\theta =\frac{\pi }{2}$is given by,

$\epsilon =\int f.dI$

Here, for a small strip, $dI=a.d\theta .\hat{\theta }$,

Putting value of f and dI , integrating the expression between limits 0 and localid="1658295437568" $\frac{\pi }{2}$

$$\begin{gathered} \mathrm{E}={\int }_0^{\frac{\mathrm{\pi }}{2}}{\mathrm{\omega \alpha B}}_0\sin \mathrm{\theta }\left(\hat{\mathrm{\varphi }}\times \hat{\mathrm{z}}\right).\left(\mathrm{a}.\mathrm{d\theta }.\hat{\mathrm{\theta }}\right) \\ \mathrm{E}={\mathrm{\omega \alpha B}}_0{\int }_0^{\frac{\mathrm{\pi }}{2}}\sin \mathrm{\theta }\left(\hat{\mathrm{\varphi }}\times \hat{\mathrm{z}}\right).\hat{\mathrm{\theta }}\mathrm{d\theta } \end{gathered}$$

Using cross-product property,

$$\begin{gathered} \hat{\mathrm{\theta }}.\left(\hat{\mathrm{\varphi }}\times \hat{\mathrm{z}}\right)=\hat{\mathrm{z}}.\left(\hat{\mathrm{\theta }}\times \hat{\mathrm{\varphi }}\right) \\ \hat{\mathrm{\theta }}.\left(\hat{\mathrm{\varphi }}\times \hat{\mathrm{z}}\right)=\hat{\mathrm{z}}.\hat{\mathrm{r}} \\ \hat{\mathrm{\theta }}.\left(\hat{\mathrm{\varphi }}\times \hat{\mathrm{z}}\right)=\mathrm{cos\theta } \end{gathered}$$

Solving expression,

$$\begin{gathered} \mathrm{E}={\mathrm{\omega \alpha }}^2{\mathrm{B}}_0{\int }_0^{\frac{\mathrm{\pi }}{2}}\sin \mathrm{\theta }\cos \mathrm{\theta }.\mathrm{d\theta } \\ \mathrm{E}={\mathrm{\omega \alpha }}^2{\mathrm{B}}_0{\left[\frac{{\sin }^2\mathrm{\theta }}{2}\right]}_0^{\frac{\mathrm{\pi }}{2}} \\ \mathrm{E}=\frac{1}{2}{\mathrm{B}}_0{\mathrm{\omega \alpha }}^2\left[{\sin }^2\frac{\mathrm{\pi }}{2}-{\sin }^{2}0\right] \\ \mathrm{E}=\frac{1}{2}{\mathrm{B}}_0{\mathrm{\omega \alpha }}^2\left[1-0\right] \end{gathered}$$

Solve further as:

$\mathrm{E}=\frac{1}{2}{\mathrm{B}}_0{\mathrm{\omega \alpha }}^2$

Hence, the emf developed between the “north pole” and the equator is$\frac{1}{2}{\mathrm{B}}_0{\mathrm{\omega \alpha }}^2$ .