Question

A conducting rod of mass m and length l is placed in the east-west direction on a frictionless horizontal table. The horizontal component of earth's magnetic field is B from south to north. If charge q is suddenly passed through the conductor, then the instantaneous velocity acquired by the rod is

• A. $\frac{2Blq}{m}$
• B. $\frac{Blq}{2m}$
• C. $\sqrt{\frac{Blq}{m}}$
• D. $\frac{(2Blq{)}^2}{m^2}$

Answer 1

The correct option is C $\sqrt{\frac{Blq}{m}}$

A conducting rod of mass$=m$

Length$=l$Magnetic field$=B$

Charge$=q$

Now the required formula of the velocity using

$\frac{1}{2}m{v}^2=qE\frac{1}{2}m{v}^2=qBl[\because F=Bl]v=\sqrt{\frac{2qBl}{m}}$

Since, Impulse$=$ Change in linear momentum

$\int F.dt=\int \left(ilB\right)dtmV\int idt=qm{v}^2=Blq{v}^2=\sqrt{\frac{Blq}{m}}$