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A triangular wire frame (each side = 2m) is placed in a region of time variant magnetic field having $dB/dt = \sqrt{3} T/s$ . The magnetic field is perpendicular to the plane of the 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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0.4 V

1.2 V

0.6 V

None

answer is A.

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Detailed Solution

V_{AB} = \frac{r}{2} \sqrt{3} (\frac{1}{2}) 2 = \frac{1}{2} (\frac{2}{\sqrt{3}}) \sqrt{3} (\frac{1}{2}) 2 = 1 V

$[r = \frac{2}{\sqrt{3}}]; | E | = \frac{r}{2} \frac{dB}{dt} = \frac{r}{2} \sqrt{3}$

Similarly in rod BC and CA induced p.d in rod AB

$V_{AB} = V_{BC} = V_{CA} = 1 V$

$V_{AB} = \int E \cos θ d l = E \cos θ \int d l = (E \cos θ) l$

Now p.d cross AB is given by

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