Write the equation for the magnetic field on | vedclass

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

Question

(D) The magnetic field $B$ on the axis of a circular loop of radius $R$ carrying current $I$ at a distance $x$ from the center is given by the formula

Vedclass · Accepted Answer

$B = \frac{\mu_0 I}{4\sqrt{2}R}$

Vedclass · Answer

(D) The magnetic field $B$ on the axis of a circular loop of radius $R$ carrying current $I$ at a distance $x$ from the center is given by the formula: $B = \frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}}$.Given that the distance from the center is equal to the radius, we have $x = R$.Substituting $x = R$ into the formula:$B = \frac{\mu_0 I R^2}{2(R^2 + R^2)^{3/2}}$$B = \frac{\mu_0 I R^2}{2(2R^2)^{3/2}}$$B = \frac{\mu_0 I R^2}{2(2^{3/2} R^3)}$$B = \frac{\mu_0 I}{2 \cdot 2^{3/2} R}$$B = \frac{\mu_0 I}{2^{5/2} R}$ or $B = \frac{\mu_0 I}{4\sqrt{2} R}$.

Write the equation for the magnetic field on the axis of a circular current-carrying loop at a distance from the center equal to its radius.

Similar Questions

In the given figure,find the magnetic field at point $O$.

$A$ thin rod is bent in the shape of a small circle of radius $r$. If the charge per unit length of the rod is $\sigma$,and if the circle is rotated about its axis at the rate of $n$ rotations per second,the magnetic induction at a point on the axis at a large distance $y$ from the centre is

$A$ cylindrical conductor of radius $R$ carries a current $i$. The value of the magnetic field at a point, which is $R/4$ distance inside from the surface, is $10 \ T$. Find the value of the magnetic field at a point, which is $4R$ distance outside from the surface.

Which law is useful to determine the relation between current and the magnetic field produced by it?

Vedclass Test Series

Exam Paper Generator

Online Exam Module