Question

A rod of length l rotates with a uniform angular velocity $\omega$ about its perpendicular bisector. A uniform magnetic field B exists parallel to the axis of rotation. The potential difference betwben the two ends of the rod is

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

Answer 1

Answer: (A)

Given $\rightarrow$ Arod of length $l$
angular velocity $\rightarrow \omega$
A magnetic field $B$ parallel to the axis of rotation.

We knaw that, $d \varepsilon=B d x \times \omega x$.
$d \varepsilon=B \omega x d x$
Integrating both sides,
$\varepsilon=\int_{-1/_{2}}^{1 / 2} B \omega x d x .$
$\varepsilon=B \omega \int\limits_{-1 / 2}^{1 / 2} x d x$
$\varepsilon=B \omega\left(\frac{x^{2}}{2}\right)_{-1 / 2}^{4 / 2} .$
$\varepsilon=\frac{B \omega\left[\frac{l^{2}}{2}-\frac{l^{2}}{4}\right]}{\varepsilon=0}$
$\varepsilon=0$
Hence potential difference between $2$ ends of the rod is zero.