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(D) The magnetic flux $\Phi$ through the loop is given by $\Phi = B A \cos 	heta$.Here,$A = (2 	ext{ cm})^2 = (0.02 	ext{ m})^2 = 4 	imes 10^{-4}

Vedclass · Accepted Answer

$24$

Vedclass · Answer

(D) The magnetic flux $\Phi$ through the loop is given by $\Phi = B A \cos 	heta$.Here,$A = (2 	ext{ cm})^2 = (0.02 	ext{ m})^2 = 4 	imes 10^{-4} 	ext{ m}^2$ and $	heta = 60^{\circ}$.Substituting the values: $\Phi = (0.4 \sin(300t)) 	imes (4 	imes 10^{-4}) 	imes \cos(60^{\circ})$.Since $\cos(60^{\circ}) = 0.5$,we get $\Phi = 0.4 	imes 4 	imes 10^{-4} 	imes 0.5 	imes \sin(300t) = 0.8 	imes 10^{-4} \sin(300t) 	ext{ Wb}$.The induced emf $\epsilon$ is given by Faraday's law: $\epsilon = -\frac{d\Phi}{dt}$.$\epsilon = -\frac{d}{dt} [0.8 	imes 10^{-4} \sin(300t)] = -0.8 	imes 10^{-4} 	imes 300 	imes \cos(300t) = -0.024 \cos(300t) 	ext{ V}$.The maximum induced emf is the amplitude of this expression,which is $\epsilon_{max} = 0.024 	ext{ V}$.Converting to millivolts: $0.024 	ext{ V} = 24 	ext{ mV}$.

$A$ square loop of side $2 \text{ cm}$ is placed in a time-varying magnetic field with magnitude $B = 0.4 \sin(300t) \text{ T}$. The normal to the plane of the loop makes an angle of $60^{\circ}$ with the field. The maximum induced emf produced in the loop is . . . . . . $\text{mV}$.

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