Question

A rod of length $l$ rotates with a uniform angular velocity $\omega$ rad/s about an axis passing through its middle point but normal to its length in a uniform magnetic field of induction $B$ with its direction parallel to the axis of rotation. The induced emf between the two ends of the rod is :

• A. $\frac{B{l}^2\omega }{2}$
• B. $zero$
• C. $\frac{B{l}^2\omega }{4}$
• D. $2B{l}^2\omega$

Answer 1

Answer: (B)

$zero$

Note that an EMF always generated along the direction $(\overrightarrow{V}\times \overrightarrow{B})$

For circular motion, $\overrightarrow{V}=\overrightarrow{R}\omega$

Hence, EMF induced, $\epsilon =\omega (\overrightarrow{R}\times \overrightarrow{B})$

Now, the two halves of the rod equal, $OA=OB$

thus, $|\overrightarrow{R_1}|=|\overrightarrow{R_2}|$

but $\overrightarrow{R_1}=-\overrightarrow{R_2}$.

Therefore, the induced EMFs will be in the opposite direction from the point $O$.

Thus two EMFs ${\epsilon }_1$ and ${\epsilon }_2$ are generated in equal and opposite direction, leading to a net ZERO potential difference across

the ends.