The magnetic induction at O due to the whole | vedclass

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

Question

(C) The conductor consists of three parts: two straight segments $AB$ and $CD$,and a semicircular arc $BC$ of radius $r$.For the straight segments $AB

Vedclass · Accepted Answer

$\frac{\mu_0 i}{4r}$

Vedclass · Answer

(C) The conductor consists of three parts: two straight segments $AB$ and $CD$,and a semicircular arc $BC$ of radius $r$.For the straight segments $AB$ and $CD$,the point $O$ lies on the axis of the current-carrying wires. According to the Biot-Savart law,the magnetic field $dB = \frac{\mu_0}{4\pi} \frac{i dl \sin 	heta}{r^2}$. For these segments,the angle $	heta$ between the current element $dl$ and the position vector is $0^\circ$ or $180^\circ$,so $\sin 	heta = 0$. Thus,the magnetic induction due to $AB$ and $CD$ is zero.For the semicircular arc $BC$,the magnetic field at the center $O$ is given by $B = \frac{1}{2} \left( \frac{\mu_0 i}{2r} ight) = \frac{\mu_0 i}{4r}$.Therefore,the total magnetic induction at $O$ is $\frac{\mu_0 i}{4r}$.

The magnetic induction at $O$ due to the whole length of the conductor is:

Similar Questions

$A$ 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 very long,straight and insulated wires are kept at a $90^o$ angle from each other in the $xy$-plane as shown in the figure. These wires carry a current of equal magnitude $I$,whose directions are shown in the figure. The net magnetic field at point $P$ will be:

Two long straight parallel conductors $A$ and $B$ carrying currents $4.5 \ A$ and $8 \ A$ respectively are separated by $25 \ cm$ in air. The resultant magnetic field at a point $P$ which is at a distance of $15 \ cm$ from conductor $A$ and $10 \ cm$ from conductor $B$ is:

Write the formula for the magnetic field at the centre of a circular current-carrying loop.

The magnetic induction at the centre of a current-carrying circular coil of radius $10 \, cm$ is $5\sqrt{5}$ times the magnetic induction at a point on its axis. The distance of the point from the centre of the coil (in $cm$) is

Vedclass Test Series

Exam Paper Generator

Online Exam Module