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A triangular wire frame (each side = 2 m) is placed in a region of time variant magnetic field $d B / d t = \sqrt{3} T / s$ . The magnetic field is perpendicular to the plane of the triangle and its centre coincides with the centre of triangle. The base of the triangle AB has a resistance $1 Ω$ while the other two sides have resistance $2 Ω$ each. The magnitude of potential difference between the points A and B will be

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1.2 V

0.6 V

None

0.4 V

answer is A.

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Detailed Solution

$e = \frac{d ϕ}{d t} = A \frac{d B}{d t} = \frac{1}{2} 2 \times 2 \sin 60^{0} \sqrt{3} = 3 V$

Induced emf in each side = e/3= 1 V

$I = \frac{e}{R} = \frac{3}{1 + 2 + 2} = \frac{3}{5} A$

$V_{A} + 1 - I \times 1 = V_{B}$

$\Rightarrow V_{A} - V_{B} = I - 1 = \frac{- 2}{5}$ = 0.4 V

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