A current-carrying loop ABCD has two circular | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The magnetic field at the centre $O$ due to a circular arc of radius $R$ subtending an angle $	heta$ (in radians) is given by $B = \frac{\mu_0 i

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(C) The magnetic field at the centre $O$ due to a circular arc of radius $R$ subtending an angle $	heta$ (in radians) is given by $B = \frac{\mu_0 i 	heta}{4 \pi R}$.Here, $	heta = 30^{\circ} = \frac{\pi}{6} 	ext{ radians}$.The straight segments $AB$ and $CD$ pass through the centre $O$, so the magnetic field due to them is zero $(B_{AB} = 0, B_{CD} = 0)$.The magnetic field due to arc $AD$ (radius $R_1 = 0.01 	ext{ m}$) is $B_{AD} = \frac{\mu_0 i}{4 \pi R_1} 	imes \frac{\pi}{6} = \frac{\mu_0 i}{24 R_1}$ (directed outwards, $\odot$).The magnetic field due to arc $BC$ (radius $R_2 = 0.02 	ext{ m}$) is $B_{BC} = \frac{\mu_0 i}{4 \pi R_2} 	imes \frac{\pi}{6} = \frac{\mu_0 i}{24 R_2}$ (directed inwards, $\otimes$).The net magnetic field is $B_{net} = B_{AD} - B_{BC} = \frac{\mu_0 i}{24} \left( \frac{1}{R_1} - \frac{1}{R_2} ight)$.Substituting the values: $B_{net} = \frac{4 \pi 	imes 10^{-7} 	imes (1.2 / \pi)}{24} \left( \frac{1}{0.01} - \frac{1}{0.02} ight) = \frac{4.8 	imes 10^{-7}}{24} (100 - 50) = 0.2 	imes 10^{-7} 	imes 50 = 10 	imes 10^{-7} 	ext{ T} = 1 \mu 	ext{T}$.

Similar Questions

In the following figure,the magnitude of the magnetic field at the point $P$ is:

$A$ straight wire of finite length carrying current $I$ subtends an angle of $60^{\circ}$ at point $P$ as shown. The magnetic field at $P$ is

Two long parallel wires are at a distance $2d$ apart. They carry steady equal currents flowing out of the plane of the paper as shown. The variation of magnetic field $B$ along the line $XX'$ is given by

$A$ conducting wire carrying current $I$ is arranged as shown in the figure. Find the magnetic field at point $O$.

Let $B_1$ be the magnitude of the magnetic field at the center of a circular coil of radius $R$ carrying current $I$. Let $B_2$ be the magnitude of the magnetic field at an axial distance $x$ from the center. For $x : R = 3 : 4$,the ratio $\frac{B_2}{B_1}$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module