Question

Figure 6.20 shows a metal rod PQ resting on the smooth rails ABand positioned between the poles of a permanent magnet. The rails,the rod, and the magnetic field are in three mutual perpendiculardirections. A galvanometer G connects the rails through a switch K.Length of the rod = 15 cm, B = 0.50 T, resistance of the closed loopcontaining the rod = 9.0 mW. Assume the field to be uniform. (a) Suppose K is open and the rod is moved with a speed of 12 cm s–1in the direction shown. Give the polarity and magnitude of theinduced emf. (b) Is there an excess charge built up at the ends of the rods whenK is open? What if K is closed? (c) With K open and the rod moving uniformly, there is no netforce 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 keepthe rod moving at the same speed (=12cm 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 fieldis parallel to the rails instead of being perpendicular?

Answer 1

(a)

The length of the given coil is 15 cm.

l=15 cm =0.15 m

Magnetic field strength, B=0.50 T.

Resistance of the loop is R=9.0 mΩ or 9× 10 −3  Ω.

The velocity at which the rod is moved is given as v=12 cm/s or 0.12 m/s

The formula to calculate the induced emf is,

e=Bvl

Substitute values in the above formula.

e=0.5×0.12×0.15 =0.009 V =9× 10 −3  V =9 mV

The Lenz’s law determines the polarity of the induced emf. The induced emf must oppose the cause of the change. With reference to the Fleming’s right hand rule, the induced current should flow from P to Q, so the end P shows positive polarity and Q shows negative polarity.

(b)

When key K is open, charge is developed at the ends of the rods and on closing the key, this charge flows in the form of current.

Therefore, yes, the excess charge is maintained due to continuous flow of the current in the circuit on closing key K.

(c)

When the rod starts to move in the magnetic field, excess change is developed at the ends of the rod which develops an electric force. This force cancels the magnetic force on the electron, thus the net force on the electrons is zero.

(d)

The formula to calculate the force on the conductor moving in magnetic field is,

F=IBl

Where,

I is the induced current.

B is the magnetic field strength.

l is the length of the conductor.

Induced current is calculated as,

I= e R = 9× 10 −3 9× 10 −3 =1 A

The length of the rod is 0.15 m.

Magnetic field strength, B=0.50 T.

Substitute the value in the above formula,

F=IBl =1×0.5×0.15 =75× 10 −3  N

Thus, the retarding force on the rod is 75× 10 −3  N.

(e)

The power from external source is required to overcome the resisting force of 75× 10 −3  N.

The velocity at which the rod moves is given as v=12 cm/s or 0.12 m/s

The power is calculated as,

P=Fv =75× 10 −3 ×0.12 =9× 10 −3  W =9 mW

When key K is open, there is no current flow in the rod, thus no power is required.

(f)

The formula to calculate the power, in terms of current i is,

Power dissipated as heat= I 2 R = ( 1 ) 2 ×9× 10 −3 =9× 10 −3 =9 mW

Thus, the power dissipated as heat is 9 mW and the source of this power is external source.

(g)

If the magnetic field is parallel to the rod, then there will not be any induced emf as the coil would not cut any flux, moving in the parallel direction.

Thus, the induced emf in the coil will be zero.