A uniform magnetic field B exists in a cylind | vedclass

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Question

(A) The square frame has a side length $a = 20 \, cm = 0.2 \, m$. One side of the square lies along the diameter of the cylindrical region of radius $

Vedclass · Accepted Answer

$3.9 	imes 10^{-5} \, A$

Vedclass · Answer

(A) The square frame has a side length $a = 20 \, cm = 0.2 \, m$. One side of the square lies along the diameter of the cylindrical region of radius $R = 10 \, cm = 0.1 \, m$. Since the side length of the square is $20 \, cm$ and the diameter of the region is $2R = 20 \, cm$,exactly half of the square frame lies within the magnetic field region. The area of the square frame inside the magnetic field is $A = \frac{1}{2} 	imes a^2 = \frac{1}{2} 	imes (0.2)^2 = 0.02 \, m^2$. The induced electromotive force $(EMF)$ is given by $|e| = A \frac{dB}{dt}$. Substituting the values: $|e| = 0.02 \, m^2 	imes 0.010 \, T/s = 2 	imes 10^{-4} \, V$. The induced current $I$ is given by $I = \frac{|e|}{R_{wire}}$. $I = \frac{2 	imes 10^{-4} \, V}{4.0 \, \Omega} = 0.5 	imes 10^{-4} \, A = 5.0 	imes 10^{-5} \, A$. Wait,re-evaluating the geometry: The area of the square inside the circle is a semicircle of radius $R$ plus a rectangle? No,the side $ab$ is the diameter. The area inside the magnetic field is a semicircle of radius $R = 10 \, cm$. Area $A = \frac{1}{2} \pi R^2 = 0.5 	imes 3.14 	imes (0.1)^2 = 0.0157 \, m^2$. $|e| = 0.0157 	imes 0.01 = 1.57 	imes 10^{-4} \, V$. $I = \frac{1.57 	imes 10^{-4}}{4} \approx 0.39 	imes 10^{-4} = 3.9 	imes 10^{-5} \, A$.

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