Question

A conducting rod of length 2l is rotating with constant singular, speed about its perpendicular bisector. A uniform magnetic field $\overrightarrow{B}$ exists parallel to the axis of rotation. The emf Induced between two ends of the rod is

A) $Bw{{l}^{2}}$
B) $\dfrac{1}{2}Bw{{l}^{2}}$
C) $\dfrac{1}{8}Bw{{l}^{2}}$
D) Zero

Answer 1

Since, a conducting rod of length 2l is rotating with constant angular speed w about its $\bot ar$ bisector and a uniform magnetic field $\overrightarrow{B}$ exists $\parallel r$ to the axis of rotation. So, we used the formula of induced emf due to rotation.

Complete answer:
Let us assume a small element dx of a distance x from the rod rotating with an angular speed w about its perpendicular bisectors. Due to the rotation, the emf induced in the small amount element is given by
$de=Bwxdx$
Where de = induced emf in the small element
Since, the rod is rotating. So the emf induced between the centre of the rod and one of its side is given by $de=\int_{0}^{1}{aw}dx$
On integrating both sides, we get
$e=Bw\left[ \dfrac{{{x}^{2}}}{2} \right]_{0}^{l}$
$e=Bw\left[ \dfrac{{{1}^{2}}-{{0}^{2}}}{2} \right]$
$e=\dfrac{1}{2}Bw{{l}^{2}}$
From the fig. $AO=OB=l$
So, the potential at rod OA ${{V}_{a}}-{{V}_{0}}=\dfrac{1}{2}Bw{{l}^{2}}\text{ }\left( 1 \right)$
And the potential at rod OB
${{V}_{B}}-{{V}_{0}}=\dfrac{1}{2}Bw{{l}^{2}}\text{ }\left( 2 \right)$
Subtracting equation 2 from 1 we get
${{V}_{A}}-{{V}_{B}}=0$

Hence, the correct option is D.

Note:
Be careful while calculating the formula induced emf $e=\dfrac{1}{2}Bw{{l}^{2}}$ and a rod of length 2l is rotating about its bisector. So its length from the centre is taken to be l and calculate the potential on each side of the rod. Use the value provided exactly at the same time.