Question

A conducting rod of length ${}^{\prime }{l}^{\prime }$ is moving with constant linear speed ${}^{\prime }{v}^{\prime }$ in a uniform magnetic field ${}^{\prime }{B}^{\prime }$. This arrangement is mutually perpendicular. Obtain the expression of motional electromotive force.

Answer 1

The emf induced in a straight conductor of length $l$ moving with velocity $v$ perpendicular to a magnetic field $B$ is,

$E=Blv$

By definition,

$E=\frac{dW}{dq}$

Magnetic field exerts a force of $F=lIB$ on a long, straight current-carrying conductor perpendicular to the field,

Distance moved in time $dt$ is, $ds=vdt$ and

The workdone is, $dW=Fds$

Putting the value of F in the equation, $dW=IlBvdt$ As we know, $I=dq/dt$,

Puttining in above equation

$dW=lBvdq$

So, motional emf is,

$E=\frac{dW}{dq}=Blv$