# The figure shows a rectangular conducting frame MNOP of resistance R placed partly in a perpendicular magnetic field $\vec{B}$ and moved with velocity $\vec{v}$ as shown in the figure.Obtain the expressions for the(a) force acting on the arm ‘ON’ and its direction, and(b) power required to move the frame to get a steady emf induced between the arms MN and PO.

(a) The induced emf in the moving conductor MNOP

$e=BlV$

where B = Magnetic Field

l= length of the conductor

The induced current, $i=\frac{e}{r}=\frac{BlV}{r}$

Hence, the force on the arm 'ON' $(F)=Bil$

On putting the value of 'i' we have force,

$(F)=\frac{B\times B\times l\times V\times l}{r}$

The force is directed in the direction opposite to Velocity of the rod (v).

(b) We know that,

$Power:P=F\times V$

Where F = Force

V = velocity
We know, $F=\frac{B^{2}l^{2}V}{r}$

therefore, $P=\frac{B^{2}l^{2}V^2}{r}$

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