Question

Figure 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.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?

Answer 1

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