A current loop, having two circular arcs join | vedclass

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(D) The magnetic field at the center of a circular arc of radius $r$ subtending an angle $	heta$ (in radians) at the center is given by $B = \frac{\m

Vedclass · Accepted Answer

$1.0 	imes 10^{-5} \, T$

Vedclass · Answer

(D) The magnetic field at the center of a circular arc of radius $r$ subtending an angle $	heta$ (in radians) at the center is given by $B = \frac{\mu_0 I 	heta}{4 \pi r}$.In the given figure, the radial segments do not contribute to the magnetic field at $O$ because the current is parallel to the position vector (or the angle between $dl$ and $r$ is $0^\circ$ or $180^\circ$).The two arcs have radii $r_1 = 3 \, cm + 2 \, cm = 5 \, cm = 0.05 \, m$ and $r_2 = 3 \, cm = 0.03 \, m$. The angle subtended is $	heta = 45^\circ = \frac{\pi}{4} \, radians$.The magnetic fields due to the two arcs are in opposite directions. The net magnetic field $B$ is:$B = B_2 - B_1 = \frac{\mu_0 I 	heta}{4 \pi r_2} - \frac{\mu_0 I 	heta}{4 \pi r_1} = \frac{\mu_0 I 	heta}{4 \pi} \left( \frac{1}{r_2} - \frac{1}{r_1} ight)$Substituting the values:$B = \frac{10^{-7} 	imes 10 	imes \frac{\pi}{4}}{1} \left( \frac{1}{0.03} - \frac{1}{0.05} ight)$$B = 10^{-6} 	imes \frac{\pi}{4} \left( \frac{5 - 3}{0.15} ight) = 10^{-6} 	imes \frac{\pi}{4} 	imes \frac{2}{0.15} = 10^{-6} 	imes \frac{\pi}{0.3} = \frac{\pi}{3} 	imes 10^{-5} \approx 1.047 	imes 10^{-5} \, T$.Thus, the value is close to $1.0 	imes 10^{-5} \, T$.

$A$ current loop, having two circular arcs joined by two radial lines, is shown in the figure. It carries a current of $10 \, A$. The magnetic field at point $O$ will be close to:

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