A rectangular conductor LMNO is placed in a uniform magnetic field of $0.5\,\mathrm{T}$, directed perpendicular to the plane of the conductor. When the arm MN of length $20\,\mathrm{cm}$ is moved towards the left with a velocity of $10\,\mathrm{m\,s^{-1}}$, calculate the emf induced in the arm. Given the resistance of the arm to be $5\,\Omega$ (other arms of negligible resistance), find the value of the current in the arm.

$\varepsilon = 1\,\mathrm{V}$; $i = 0.2\,\mathrm{A}$.

CBSE Class 12 · 2013 · Delhi · Electromagnetic InductionShort Answer

A rectangular conductor LMNO is placed in a uniform magnetic field of

0.5 T

, directed perpendicular to the plane of the conductor. When the arm MN of length

20 cm

is moved towards the left with a velocity of

10 m s^{- 1}

, calculate the emf induced in the arm. Given the resistance of the arm to be

5 Ω

(other arms of negligible resistance), find the value of the current in the arm.

Answer: $ε = 1 V$ ; $i = 0.2 A$ .

Solution

The arm MN cuts field lines as it moves, producing a motional emf

ε = B ℓ v = 0.5 \times 0.20 \times 10 = 1 V

The induced current is

i = \frac{ε}{R} = \frac{1}{5} = 0.2 A

ε = 1 V, i = 0.2 A

CBSE Class 12 Board · 2013 · Delhi · Subjective

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