A circular loop of radius r is carrying curre | vedclass

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

Question

(C) The magnetic field $B$ at a distance $x$ from the centre of a circular loop of radius $r$ carrying current $I$ on its axis is given by:$B = \frac{

Vedclass · Accepted Answer

$2\sqrt{2} : 1$

Vedclass · Answer

(C) The magnetic field $B$ at a distance $x$ from the centre of a circular loop of radius $r$ carrying current $I$ on its axis is given by:$B = \frac{\mu_0 I r^2}{2(r^2 + x^2)^{3/2}}$At the centre of the loop,$x = 0$:$B_1 = \frac{\mu_0 I r^2}{2(r^2 + 0)^{3/2}} = \frac{\mu_0 I r^2}{2r^3} = \frac{\mu_0 I}{2r}$At a distance $x = r$ on the axis:$B_2 = \frac{\mu_0 I r^2}{2(r^2 + r^2)^{3/2}} = \frac{\mu_0 I r^2}{2(2r^2)^{3/2}} = \frac{\mu_0 I r^2}{2(2^{3/2} r^3)} = \frac{\mu_0 I}{2(2\sqrt{2}r)} = \frac{\mu_0 I}{4\sqrt{2}r}$The ratio of the magnetic field at the centre to the magnetic field at distance $r$ is:$\frac{B_1}{B_2} = \frac{\frac{\mu_0 I}{2r}}{\frac{\mu_0 I}{4\sqrt{2}r}} = \frac{4\sqrt{2}}{2} = 2\sqrt{2}$Thus,the ratio is $2\sqrt{2} : 1$.

$A$ circular loop of radius $r$ is carrying current $I \ A$. The ratio of the magnetic field at the centre of the circular loop to the magnetic field at a distance $r$ from the centre of the loop on its axis is:

Similar Questions

Write the equation for the magnetic field of a current-carrying circular loop at: $(i)$ a point on the axis,and $(ii)$ a point on the plane of the loop at a distance $x$ from the center of the loop.

$A$ circular coil of wire consisting of $100$ turns,each of radius $8.0 \; cm$,carries a current of $0.40 \; A$. What is the magnitude of the magnetic field $B$ at the centre of the coil?

$A$ circular loop of radius $0.0157\,m$ carries a current of $2.0\,A$. The magnetic field at the centre of the loop is $(\mu_0 = 4\pi \times 10^{-7}\,T\cdot m/A)$.

Two identical long current-carrying wires are bent into the shapes shown in the following figures. If the magnitudes of the magnetic fields at the centers $P$ and $Q$ of the semicircular arcs are $B_1$ and $B_2$ respectively,then the ratio $\frac{B_1}{B_2}$ is . . . . . . .

An equilateral triangle of side '$a$' carries a current '$i$'. What is the magnetic field at point '$P$' (a vertex of the triangle)?

Vedclass Test Series

Exam Paper Generator

Online Exam Module