Figure 6.20 shows a metal rod PQ resting on the smooth rails AB and positioned between the poles of a permanent magnet. The rails, the rod, and the magnetic field are in three mutual perpendicular directions. A galvanometer G connects the rails through a switch K. Length of the rod = 15 cm, *B *= 0.50 T, resistance of the closed loop containing the rod = 9.0 mΩ. Assume the field to be uniform.

**(a) **Suppose K is open and the rod is moved with a speed of 12 cm s^{−1} in the direction shown. Give the polarity and magnitude of the induced emf.

**(b) **Is there an excess charge built up at the ends of the rods when

K is open? What if K is closed?

**(c) **With K open and the rod moving uniformly, there is *no net force *on the electrons in the rod PQ even though they do experience magnetic force due to the motion of the rod. Explain.

**(d) **What is the retarding force on the rod when K is closed?

**(e) **How much power is required (by an external agent) to keep the rod moving at the same speed (=12 cm s^{−1}) when K is closed? How much power is required when K is open?

**(f) **How much power is dissipated as heat in the closed circuit?

What is the source of this power?

**(g) **What is the induced emf in the moving rod if the magnetic field is parallel to the rails instead of being perpendicular?

#### Solution

Length of the rod,* l* = 15 cm = 0.15 m

Magnetic field strength,* B* = 0.50 T

Resistance of the closed loop, *R* = 9 mΩ = 9 × 10^{−3} Ω

**(a)** Induced emf = 9 mV; polarity of the induced emf is such that end *P* shows positive while end *Q*shows negative ends.

Speed of the rod, *v* = 12 cm/s = 0.12 m/s

Induced emf is given as:

*e* = *Bvl*

*= *0.5 × 0.12 × 0.15

= 9 × 10^{−3} v

= 9 mV

The polarity of the induced emf is such that end *P* shows positive while end *Q* shows negative ends.

**(b) **Yes; when key K is closed, excess charge is maintained by the continuous flow of current.

When key K is open, there is excess charge built up at both ends of the rods.

When key K is closed, excess charge is maintained by the continuous flow of current.

**(c) **Magnetic force is cancelled by the electric force set-up due to the excess charge of opposite nature at both ends of the rod.

There is no net force on the electrons in rod PQ when key K is open and the rod is moving uniformly. This is because magnetic force is cancelled by the electric force set-up due to the excess charge of opposite nature at both ends of the rods.

**(d) **Retarding force exerted on the rod, *F* = *IBl*

Where,

*I* = Current flowing through the rod

`=e/R=(9xx10^-3)/(9xx10^-3)=1A`

`F=1xx0.5xx0.15`

`=75xx10^-3N`

**(e)** 9 mW; no power is expended when key K is open.

Speed of the rod, *v* = 12 cm/s = 0.12 m/s

Hence, power is given as:

`P=Fv`

`=75xx10^-3xx0.12`

`=9xx10^-3 W`

`=9 mW`

When key K is open, no power is expended.

**(f)** 9 mW; power is provided by an external agent.

Power dissipated as heat = *I*^{2} *R*

= (1)^{2} × 9 × 10^{−3}

= 9 mW

The source of this power is an external agent.

**(g)** Zero

In this case, no emf is induced in the coil because the motion of the rod does not cut across the field lines.