A circular loop of radius 20 text{ cm} and re | vedclass

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Question

(A) The magnetic flux $\phi$ through the loop is given by $\phi = B \cdot A = (2t^2 + 2t + 3) \cdot \pi r^2$.Given $r = 20 	ext{ cm} = 0.2 	ext{ m}$

Vedclass · Accepted Answer

$44$

Vedclass · Answer

(A) The magnetic flux $\phi$ through the loop is given by $\phi = B \cdot A = (2t^2 + 2t + 3) \cdot \pi r^2$.Given $r = 20 	ext{ cm} = 0.2 	ext{ m}$, so $A = \pi (0.2)^2 = 0.04\pi 	ext{ m}^2$.Thus, $\phi = 0.04\pi (2t^2 + 2t + 3) 	ext{ Wb}$.The induced emf $\varepsilon$ is given by Faraday's law: $\varepsilon = |-\frac{d\phi}{dt}|$.$\frac{d\phi}{dt} = 0.04\pi (4t + 2)$.At $t = 3 	ext{ s}$, $\frac{d\phi}{dt} = 0.04\pi (4(3) + 2) = 0.04\pi (14) = 0.56\pi 	ext{ V}$.The induced current $I$ is $I = \frac{\varepsilon}{R} = \frac{0.56\pi}{2} = 0.28\pi 	ext{ A}$.Using $\pi \approx \frac{22}{7}$, $I = 0.28 	imes \frac{22}{7} = 0.04 	imes 22 = 0.88 	ext{ A}$.Given $I = \frac{\alpha}{50}$, we have $0.88 = \frac{\alpha}{50}$.Therefore, $\alpha = 0.88 	imes 50 = 44$.

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