A wire loop enclosing a semi-circle of radius | vedclass

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Question

(A) The magnetic flux $\phi_B$ through the part of the loop inside the magnetic field is given by the area of the sector of the circle within the fiel

Vedclass · Accepted Answer

$\frac{B R^2}{2} \omega$

Vedclass · Answer

(A) The magnetic flux $\phi_B$ through the part of the loop inside the magnetic field is given by the area of the sector of the circle within the field multiplied by the magnetic field strength $B$.Let $	heta$ be the angle of the sector inside the magnetic field at any time $t$. The area of this sector is $A = \frac{1}{2} R^2 	heta$.The magnetic flux is $\phi_B = B A = B \left( \frac{1}{2} R^2 	heta ight)$.According to Faraday's law of induction,the magnitude of the induced emf $\varepsilon$ is given by $\varepsilon = \left| \frac{d \phi_B}{dt} ight|$.Since the loop rotates with angular velocity $\omega$,we have $\frac{d 	heta}{dt} = \omega$.Therefore,$\varepsilon = \left| \frac{d}{dt} \left( \frac{1}{2} B R^2 	heta ight) ight| = \frac{1}{2} B R^2 \left| \frac{d 	heta}{dt} ight| = \frac{B R^2 \omega}{2}$.

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