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 dissipated as heat in the closed circuit? What is the source of this power?

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

Answer 2

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Thanks , For Your Question ...

=) Given ,
Length of the rod (l)  = 15 cm
Converting it into meter = 0.15 m

Magnetic field strength (B)  = 0.50 T

Resistance of the closed loop (R)  = 9 mΩ
So , It will be , = 9 × 10^-3 Ω

To find :- Power dissipated as heat in the closed circuit .

we know , Power dissipated = I^2 R
=) (1)^2 × (9 × 10^-3)
=) 9 × 10^-3 W (answer)

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Q.) What is the source of this power ?

Ans.) External agent is the source of this power .

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