A circular coil of radius R is carrying a cur | vedclass

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

Question

(C) The magnetic field at the centre of the circular coil due to current $I_{1}$ is given by: $B_{1} = \frac{\mu_{0} I_{1}}{2 R}$ The magnetic field a

Vedclass · Accepted Answer

$\frac{R}{\pi}\left(\frac{I_{2}}{I_{1}}\right)$

Vedclass · Answer

(C) The magnetic field at the centre of the circular coil due to current $I_{1}$ is given by: $B_{1} = \frac{\mu_{0} I_{1}}{2 R}$ The magnetic field at the centre of the coil due to the long straight wire carrying current $I_{2}$ at a distance $d$ is given by: $B_{2} = \frac{\mu_{0} I_{2}}{2 \pi d}$ For the net magnetic field at the centre to be zero,the magnitudes of these fields must be equal and their directions must be opposite: $B_{1} = B_{2}$ Substituting the expressions: $\frac{\mu_{0} I_{1}}{2 R} = \frac{\mu_{0} I_{2}}{2 \pi d}$ Solving for $d$: $d = \frac{\mu_{0} I_{2}}{2 \pi} 	imes \frac{2 R}{\mu_{0} I_{1}}$ $d = \frac{I_{2} R}{\pi I_{1}}$ Thus,the correct option is $C$.

Similar Questions

Write the formula for the magnetic field at a perpendicular distance $r$ from a finite current-carrying wire.

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:

Two circular coils $P$ and $Q$ are made from two identical wires of the same length. The number of turns in coils $P$ and $Q$ are $4$ and $2$,respectively. The magnetic inductions at the centers of $P$ and $Q$ are $B_P$ and $B_Q$,respectively. The ratio $\frac{B_P}{B_Q}$ is

For the adjoining figure,the magnetic field at point $P$ will be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module