Question

A metallic rod PQ of length l is rotated with an angular velocity o about
an axis passing through its mid-pint (0) and perpendicular to the plane
of the paper, in uniform magnetic field B , as shown in the figure. What
is the potential difference developed between the two ends of the rod, P
and Q?​

Answer 1

Given:

A metallic rod PQ of length l is rotated with an angular velocity $\omega$ about an axis passing through its mid-pint (0) and perpendicular to the plane of the paper, in uniform magnetic field B.

To find:

Net potential difference develop between the two ends of the rod.

Calculation:

OQ segment of wire has length l/2.

So, potential difference generated by OQ segment of wire is given as:

$E1 = vB( \dfrac{l}{2} ) \sin( {90}^{ \circ} )$

$= > E1 = \dfrac{vBl}{2}$

$= > E1 = \dfrac{( \omega l)Bl}{2}$

$= > E1 = \dfrac{\omega B {l}^{2} }{2}$

Similarly PO segment has length l/2.

So, potential difference generated by PO segment of the wire is given as:

$E2= vB( \dfrac{l}{2} ) \sin( {90}^{ \circ} )$

$= > E2= \dfrac{vBl}{2}$

$= > E2 = \dfrac{( \omega l)Bl}{2}$

$= > E2 = \dfrac{\omega B {l}^{2} }{2}$

The Potential Difference between the 2 ends of rod is :

$= > E2 - E1= \dfrac{\omega B {l}^{2} }{2} - \dfrac{\omega B {l}^{2} }{2}$

$= > E2 - E1= 0 \: volt$

So, final answer is:

∆E = 0 volt.