The magnetic field at the centre of a current | 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 a circular coil of radius $R$ carrying current $I$ is given by $B_c = \frac{\mu_0 I}{2R}$.The magnetic field a

Vedclass · Accepted Answer

$2 \sqrt{2}$

Vedclass · Answer

(C) The magnetic field at the centre of a circular coil of radius $R$ carrying current $I$ is given by $B_c = \frac{\mu_0 I}{2R}$.The magnetic field at a point on the axis of the coil at a distance $x$ from the centre is given by $B_a = \frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}}$.Given $x = R$,we substitute this into the formula for $B_a$:$B_a = \frac{\mu_0 I R^2}{2(R^2 + R^2)^{3/2}} = \frac{\mu_0 I R^2}{2(2R^2)^{3/2}} = \frac{\mu_0 I R^2}{2(2^{3/2} R^3)} = \frac{\mu_0 I}{2 \cdot 2 \sqrt{2} R} = \frac{\mu_0 I}{4 \sqrt{2} R}$.Now,we find the ratio $\frac{B_c}{B_a}$:$\frac{B_c}{B_a} = \frac{\frac{\mu_0 I}{2R}}{\frac{\mu_0 I}{4 \sqrt{2} R}} = \frac{4 \sqrt{2}}{2} = 2 \sqrt{2}$.Therefore,the correct option is $C$.

The magnetic field at the centre of a current-carrying circular coil of radius $R$ is $B_c$ and the magnetic field at a point on its axis at a distance $R$ from its centre is $B_a$. The value of $\frac{B_c}{B_a}$ is

Similar Questions

An element $\overrightarrow{\Delta \ell} = \Delta x \hat{i}$ is placed at the origin and carries a current of $10 \ A$. Find the magnitude of the magnetic field on the $Y$-axis at a distance of $0.5 \ m$,given $\Delta x = 1 \ cm$. (Use $\frac{\mu_0}{4 \pi} = 10^{-7} \ T \cdot m/A$)

Two identical coils having the same number of turns and carrying equal current have a common centre,and their planes are at right angles to each other. What is the ratio of the magnitude of the resultant magnetic field at the centre to the magnetic field due to one of the coils at the centre?

The Biot-Savart law for a moving point charge $q$ with velocity $\vec{v}$ is given by $\vec{B} = \frac{\mu_0}{4\pi} \frac{q(\vec{v} \times \vec{r})}{r^3}$. This indicates that the magnetic field $\vec{B}$ produced by an electron moving with velocity $\vec{v}$ is such that:

$A$ battery is connected between two points $A$ and $B$ on the circumference of a uniform conducting ring of radius $r$ and resistance $R$. One of the arcs $AB$ of the ring subtends an angle $\theta$ at the centre. The value of the magnetic induction at the centre due to the current in the ring is

In the given diagram,two current-carrying circular loops of radius $R$ and $2R$ are arranged in the $YZ-$ plane and $XZ-$ plane respectively. The common center of both is at the origin $O$. What will be the angle of the resultant magnetic field from the $X-$ axis?

Vedclass Test Series

Exam Paper Generator

Online Exam Module