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(B) The induced $EMF$ in a rotating loop is given by $E = B A \omega \sin(	heta)$, where $	heta = \omega t$ is the angle between the area vector and

Vedclass · Accepted Answer

$14$

Vedclass · Answer

(B) The induced $EMF$ in a rotating loop is given by $E = B A \omega \sin(	heta)$, where $	heta = \omega t$ is the angle between the area vector and the magnetic field. Given: $B = 0.5 \ T$, $\omega = 100 \ rad/s$, $E = 15.4 \ mV = 15.4 	imes 10^{-3} \ V$, and $	heta = 30^{\circ}$. The area of the circular loop is $A = \pi r^2$. Substituting the values into the formula: $15.4 	imes 10^{-3} = 0.5 	imes (\pi r^2) 	imes 100 	imes \sin(30^{\circ})$. Since $\sin(30^{\circ}) = 0.5$, we have: $15.4 	imes 10^{-3} = 0.5 	imes \pi r^2 	imes 100 	imes 0.5$. $15.4 	imes 10^{-3} = 25 	imes \pi r^2$. Using $\pi = 22/7$: $15.4 	imes 10^{-3} = 25 	imes (22/7) 	imes r^2$. $r^2 = \frac{15.4 	imes 10^{-3} 	imes 7}{25 	imes 22} = \frac{107.8 	imes 10^{-3}}{550} = 0.196 	imes 10^{-3} = 196 	imes 10^{-6} \ m^2$. $r = \sqrt{196 	imes 10^{-6}} = 14 	imes 10^{-3} \ m = 14 \ mm$.

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