A $0.5\,\mathrm{m}$ long metal rod PQ completes the circuit as shown in the figure. The plane of the circuit is perpendicular to a magnetic field of flux density $0.15\,\mathrm{T}$. If the resistance of the total circuit is $3\,\Omega$, calculate the force needed to move the rod in the direction indicated with a constant speed of $2\,\mathrm{m\,s^{-1}}$.

$F = 3.75 \times 10^{-3}\,\mathrm{N}$.

CBSE Class 12 · 2006 · Electromagnetic InductionShort Answer

0.5 m

long metal rod PQ completes the circuit as shown in the figure. The plane of the circuit is perpendicular to a magnetic field of flux density

0.15 T

. If the resistance of the total circuit is

3 Ω

, calculate the force needed to move the rod in the direction indicated with a constant speed of

2 m s^{- 1}

Answer: $F = 3.75 \times 1 0^{- 3} N$ .

Solution

The motional emf is

ε = B ℓ v

, driving a current

i = \frac{B ℓ v}{R}

through the circuit.
To keep the rod moving at constant speed, the applied force must balance the magnetic force

F = B i ℓ

on the current-carrying rod:

F = B i ℓ = \frac{B ^{2} ℓ ^{2} v}{R} = \frac{( 0.15 ) ^{2} ( 0.5 ) ^{2} ( 2 )}{3}

F = 3.75 \times 1 0^{- 3} N

CBSE Class 12 Board · 2006 · Subjective

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