Question

A conducting rod of length $2l$ is rotating with constant angular speed $\omega$ about its perpendicular bisector. A uniform magnetic field $B^{\to }$ exists parallel to the axis of rotation. The e.m.f., induced between two ends of the rod is

• A. $\frac{1}{2}Bl{\omega }^2$
• B. $Bl{\omega }^2$
• C. Zero
• D. $\frac{1}{8}Bl{\omega }^2$

Answer 1

The correct option is C

Zero

Step 1:Given:

Length of rod =2l

Angular speed = $\omega$

Magnetic field =B

Step 3 : Finding $\epsilon$(EMF) induced between 2 ends of rod

1. EMF- Electromotive force of a rod in a magnetic field is equal to the product of its length, Magnetic strength, and linear velocity.
2. $\epsilon =BvL$

To find emf induced between 2 ends of a rod we begin with finding emf induced in a small piece (dx) of the rod at a distance of x from the center of the rod.

$$\begin{gathered} Linearspeedofthissmallpiecewillbeintermsof\omega \\ v=\omega x \\ LengthisdxandthemagneticfieldisB \\ d\epsilon =Bdx\times \omega x \end{gathered}$$

To get emf of the whole rod, we need to integrate $d\epsilon$from $-lto+l$

$$\begin{gathered} \int d\epsilon ={\int }_{-l}^{+l}Bdx\omega x \\ \int d\epsilon =B\omega {\int }_{-l}^{+l}x.dx \\ \epsilon =B\omega {[\frac{x^2}{2}]}_{-l}^l\epsilon =B\omega [\frac{l^2}{2}-\frac{l^2}{2}] \\ \epsilon =0 \end{gathered}$$

Hence, option C is right.