Plane figures made of thin wires of resistanc | vedclass

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(A) The induced electromotive force $(EMF)$ is given by $\varepsilon = -A \frac{dB}{dt}$. Since the magnetic field is decreasing, $dB/dt = -0.1 	ext{

Vedclass · Accepted Answer

$10^{-4} 	ext{ A}$ (Clockwise), $2 	imes 10^{-4} 	ext{ A}$ (Clockwise)

Vedclass · Answer

(A) The induced electromotive force $(EMF)$ is given by $\varepsilon = -A \frac{dB}{dt}$. Since the magnetic field is decreasing, $dB/dt = -0.1 	ext{ mT/s} = -10^{-4} 	ext{ T/s}$.For the inner loop of radius $a = 0.1 	ext{ m}$:The resistance is $R_1 = (50 	imes 10^{-3} \Omega/	ext{m}) 	imes (2\pi a) = 50 	imes 10^{-3} 	imes 2\pi 	imes 0.1 = 0.01\pi \Omega$.The area is $A_1 = \pi a^2 = \pi (0.1)^2 = 0.01\pi 	ext{ m}^2$.The magnitude of induced current is $i_1 = \frac{|\varepsilon_1|}{R_1} = \frac{A_1 |dB/dt|}{R_1} = \frac{0.01\pi 	imes 10^{-4}}{0.01\pi} = 10^{-4} 	ext{ A}$.Since the magnetic field is decreasing, the induced current will try to oppose this change by creating a magnetic field in the same direction, which is clockwise.For the outer loop of radius $b = 0.2 	ext{ m}$:The resistance is $R_2 = (50 	imes 10^{-3} \Omega/	ext{m}) 	imes (2\pi b) = 50 	imes 10^{-3} 	imes 2\pi 	imes 0.2 = 0.02\pi \Omega$.The area is $A_2 = \pi b^2 = \pi (0.2)^2 = 0.04\pi 	ext{ m}^2$.The magnitude of induced current is $i_2 = \frac{|\varepsilon_2|}{R_2} = \frac{A_2 |dB/dt|}{R_2} = \frac{0.04\pi 	imes 10^{-4}}{0.02\pi} = 2 	imes 10^{-4} 	ext{ A}$.Similarly, the direction is clockwise.

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