Question

A copper rod of length L is rotating about the midpoint of rod perpendicular to the magnetic field B with constant angular velocity $\omega $. The induced EMF between the two ends is
A) $\dfrac{1}{2}B\omega {l^2}$
B) $B\omega {l^2}$
C) $2B\omega {l^2}$
D) Zero

Answer 1

Hint
When the copper rod of length is rotating perpendicularly in the magnetic field then no emf will induce in the rod. Use the concept of integration in the formula ${\rm E} = Blv$ and put the limits from $\dfrac{{ - L}}{2}\,to\,\dfrac{L}{2}$ to solve the problem.

Complete step by step answer
The copper rod is rotating about its midpoint, so we'll take the midpoint as reference and then consider a differential length of the rod dx from its centre. Since the rod is rotating with an angular velocity w and we're considering a differential area so the angular velocity of the differential area will be $\omega \times x\;$ placed in a magnetic field B. So integrating for the evaluation of EMF induced in the coil;
Hence
$\Rightarrow {\rm E} = Blv $
$\Rightarrow E = B\omega xdx$
(l is taken as dx due to the length. And v = $\omega \times x\;$)
Integrating within the given limits we have,
$\Rightarrow E = \int\limits_{\dfrac{{ - L}}{2}}^{\dfrac{L}{2}} {B\omega \,xdx} $
$\Rightarrow E = B\omega \int\limits_{\dfrac{{ - L}}{2}}^{\dfrac{L}{2}} {\,xdx} $
$\Rightarrow E = B\omega \,[\dfrac{{{x^2}}}{2}]_{ - L/2}^{L/2}$
On further solving we have,
$\Rightarrow E = B\omega \,\,\,[\dfrac{{{L^2}}}{2} - \dfrac{{{{( - L)}^2}}}{2}] $
$\Rightarrow E = 0$
Since the emf of the rod is zero and hence the correct answer is option (D).

Note
At particular times in this kind of question we are often asked to calculate the EMF from one end. That time we will put the lower limit as 0. And hence the question will be solved and the answer will not come out to be 0.