Q47P - Introduction to Electrodynamics

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:$\left[\frac{1}{2}{B}_0{\mathrm{\omega \alpha }}^2\right]$.

Step-by-Step Solution

Answer

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

Step 1: Given information

The radius of spherical shell is, a .

The spherical shell rotates about the z axis.

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

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

Step 2: 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.

Step 3: 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 $\frac{\pi }{2}$

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

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