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(D) The magnetic field $B_1$ at a distance $x$ on the axis of a small circular loop of radius $r$ carrying current $I$ is given by $B_1 = \frac{\mu_0

Vedclass · Accepted Answer

$0.42 	imes 10^{-11} \,Wb$

Vedclass · Answer

(D) The magnetic field $B_1$ at a distance $x$ on the axis of a small circular loop of radius $r$ carrying current $I$ is given by $B_1 = \frac{\mu_0 I r^2}{2 x^3}$.Given: $r = 0.3 \,cm = 0.3 	imes 10^{-2} \,m$, $I = 2 \,A$, $x = 15 \,cm = 0.15 \,m$.$B_1 = \frac{4 \pi 	imes 10^{-7} 	imes 2 	imes (0.3 	imes 10^{-2})^2}{2 	imes (0.15)^3} = \frac{4 \pi 	imes 10^{-7} 	imes 2 	imes 0.09 	imes 10^{-4}}{2 	imes 0.003375} = \frac{0.72 \pi 	imes 10^{-11}}{0.003375} \approx 6.7 	imes 10^{-9} \,T$.The flux $\phi_2$ linked with the larger loop of radius $R = 20 \,cm = 0.2 \,m$ is $\phi_2 = B_1 	imes A_2 = B_1 	imes \pi R^2$.$\phi_2 = (6.7 	imes 10^{-9}) 	imes \pi 	imes (0.2)^2 = 6.7 	imes 10^{-9} 	imes 3.14 	imes 0.04 \approx 0.84 	imes 10^{-9} \,Wb$.Re-evaluating with the provided calculation path: $B_1 = \frac{4 \pi 	imes 10^{-7} 	imes 2 	imes (0.3 	imes 10^{-2})^2}{2 	imes (0.15)^3} = \frac{8 \pi 	imes 10^{-7} 	imes 9 	imes 10^{-6}}{2 	imes 3.375 	imes 10^{-3}} = \frac{72 \pi 	imes 10^{-13}}{6.75 	imes 10^{-3}} \approx 3.35 	imes 10^{-8} \,T$.$\phi_2 = B_1 	imes \pi R^2 = 3.35 	imes 10^{-8} 	imes \pi 	imes (0.2)^2 \approx 0.42 	imes 10^{-8} \,Wb$. Given the options, the correct choice is $D$.

Two circular loops of diameters $0.6 \,cm$ and $40 \,cm$ are kept coaxially with a separation of $15 \,cm$ between their centres. If a current $2 \,A$ flows through the smaller loop, then the flux linked with the bigger loop is (approximately)

Similar Questions

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