A triangular wire frame (each side = 2 , m) i | vedclass

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(A) The area of the equilateral triangle is $A = \frac{\sqrt{3}}{4} 	imes (	ext{side})^2 = \frac{\sqrt{3}}{4} 	imes (2)^2 = \sqrt{3} \, m^2$.Accord

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$0.4$

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(A) The area of the equilateral triangle is $A = \frac{\sqrt{3}}{4} 	imes (	ext{side})^2 = \frac{\sqrt{3}}{4} 	imes (2)^2 = \sqrt{3} \, m^2$.According to Faraday's law,the induced emf is $\varepsilon = \left| \frac{d\phi}{dt} ight| = A \frac{dB}{dt} = \sqrt{3} 	imes \sqrt{3} = 3 \, V$.Since the triangle is symmetric,the induced emf in each side is $\varepsilon_{side} = \frac{\varepsilon}{3} = \frac{3}{3} = 1 \, V$.The total resistance of the circuit is $R_{total} = 1 + 2 + 2 = 5 \, \Omega$.The current in the loop is $I = \frac{\varepsilon}{R_{total}} = \frac{3}{5} = 0.6 \, A$.The potential difference between points $A$ and $B$ is given by $V_A - V_B = \varepsilon_{AB} - I 	imes R_{AB}$,where $\varepsilon_{AB} = 1 \, V$ is the induced emf in side $AB$ and $R_{AB} = 1 \, \Omega$.$V_A - V_B = 1 - (0.6 	imes 1) = 0.4 \, V$.Thus,the magnitude of the potential difference is $0.4 \, V$.

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