91影视

The radius ofspherical shellis, a .

The spherical shell rotates about the z axis.

The angular velocity of rotation is, $蝇$.

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

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 鈥榤agnetic force鈥�.

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

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

$\mathrm{v}=\mathrm{蝇伪sin}\mathrm{胃}\widehat{\mathrm{蠒}}$.

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

$$\begin{gathered} \mathrm{f}=\mathrm{v}脳\mathrm{B} \\ \mathrm{f}=\mathrm{蝇伪}\sin \mathrm{胃}\widehat{\mathrm{蠒}}脳{\mathrm{B}}_0\hat{\mathrm{z}} \\ \mathrm{f}={\mathrm{蝇伪B}}_0\sin \mathrm{胃}\left(\widehat{\mathrm{蠒}}脳\hat{\mathrm{z}}\right) \end{gathered}$$

Then the formula for the emf developed between the 鈥渘orth pole鈥�$胃=0$and the equator $胃=\frac{蟺}{2}$is given by,

$蔚=鈭�f.dI$

Here, for a small strip, $dI=a.d胃.\widehat{胃}$,

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

$$\begin{gathered} \mathrm{E}={鈭�}_0^{\frac{\mathrm{蟺}}{2}}{\mathrm{蝇伪B}}_0\sin \mathrm{胃}\left(\widehat{\mathrm{蠒}}脳\hat{\mathrm{z}}\right).\left(\mathrm{a}.\mathrm{诲胃}.\widehat{\mathrm{胃}}\right) \\ \mathrm{E}={\mathrm{蝇伪B}}_0{鈭�}_0^{\frac{\mathrm{蟺}}{2}}\sin \mathrm{胃}\left(\widehat{\mathrm{蠒}}脳\hat{\mathrm{z}}\right).\widehat{\mathrm{胃}}\mathrm{诲胃} \end{gathered}$$

Using cross-product property,

$$\begin{gathered} \widehat{\mathrm{胃}}.\left(\widehat{\mathrm{蠒}}脳\hat{\mathrm{z}}\right)=\hat{\mathrm{z}}.\left(\widehat{\mathrm{胃}}脳\widehat{\mathrm{蠒}}\right) \\ \widehat{\mathrm{胃}}.\left(\widehat{\mathrm{蠒}}脳\hat{\mathrm{z}}\right)=\hat{\mathrm{z}}.\hat{\mathrm{r}} \\ \widehat{\mathrm{胃}}.\left(\widehat{\mathrm{蠒}}脳\hat{\mathrm{z}}\right)=\mathrm{肠辞蝉胃} \end{gathered}$$

Solving expression,

$$\begin{gathered} \mathrm{E}={\mathrm{蝇伪}}^2{\mathrm{B}}_0{鈭�}_0^{\frac{\mathrm{蟺}}{2}}\sin \mathrm{胃}\cos \mathrm{胃}.\mathrm{诲胃} \\ \mathrm{E}={\mathrm{蝇伪}}^2{\mathrm{B}}_0{\left[\frac{{\sin }^2\mathrm{胃}}{2}\right]}_0^{\frac{\mathrm{蟺}}{2}} \\ \mathrm{E}=\frac{1}{2}{\mathrm{B}}_0{\mathrm{蝇伪}}^2\left[{\sin }^2\frac{\mathrm{蟺}}{2}-{\sin }^{2}0\right] \\ \mathrm{E}=\frac{1}{2}{\mathrm{B}}_0{\mathrm{蝇伪}}^2\left[1-0\right] \end{gathered}$$

Solve further as:

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

Hence, the emf developed between the 鈥渘orth pole鈥� and the equator is$\frac{1}{2}{\mathrm{B}}_0{\mathrm{蝇伪}}^2$ .