A uniform magnetic field of strength B=2 text | vedclass

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Question

(A) Given,$B = 2 	ext{ mT} = 2 	imes 10^{-3} 	ext{ T}$.Radius $R = \frac{10}{\sqrt{\pi}} 	ext{ cm} = \frac{10}{\sqrt{\pi}} 	imes 10^{-2} 	ext{ m

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$\Phi_{S_1}=-20 \mu 	ext{ Wb}, \Phi_{S_2}=+20 \mu 	ext{ Wb}$

Vedclass · Answer

(A) Given,$B = 2 	ext{ mT} = 2 	imes 10^{-3} 	ext{ T}$.Radius $R = \frac{10}{\sqrt{\pi}} 	ext{ cm} = \frac{10}{\sqrt{\pi}} 	imes 10^{-2} 	ext{ m} = \frac{10^{-1}}{\sqrt{\pi}} 	ext{ m}$.The area of the circular cross-section $S_3$ is $A = \pi R^2 = \pi 	imes \left(\frac{10^{-1}}{\sqrt{\pi}}ight)^2 = \pi 	imes \frac{10^{-2}}{\pi} = 10^{-2} 	ext{ m}^2$.For the hemisphere $S_1$,the magnetic field lines enter the surface. The angle between the area vector (outward normal) and the magnetic field vector is $180^{\circ}$.Flux $\Phi_{S_1} = B A \cos(180^{\circ}) = (2 	imes 10^{-3} 	ext{ T}) 	imes (10^{-2} 	ext{ m}^2) 	imes (-1) = -2 	imes 10^{-5} 	ext{ Wb} = -20 	imes 10^{-6} 	ext{ Wb} = -20 \mu 	ext{ Wb}$.Since the total magnetic flux through a closed surface is zero (Gauss's Law for magnetism),the flux leaving the cone $S_2$ must be equal in magnitude to the flux entering the hemisphere $S_1$.Therefore,$\Phi_{S_1} + \Phi_{S_2} = 0 \implies \Phi_{S_2} = -\Phi_{S_1} = +20 \mu 	ext{ Wb}$.

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