A circular loop of radius r of conducting wir | vedclass

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

Question

(B) The magnetic field at the centre of a circular loop of radius $r$ carrying current $I$ is given by $B = \frac{\mu_0 I}{2r}$.Case $(I)$: For a loop

Vedclass · Accepted Answer

$\frac{B}{4}$

Vedclass · Answer

(B) The magnetic field at the centre of a circular loop of radius $r$ carrying current $I$ is given by $B = \frac{\mu_0 I}{2r}$.Case $(I)$: For a loop of radius $r$,the length is $l_1 = 2\pi r$. The resistance is $R_1 = \rho \frac{l_1}{A} = \rho \frac{2\pi r}{A}$. The current is $I_1 = \frac{V}{R_1}$. The magnetic field is $B_1 = \frac{\mu_0 I_1}{2r}$.Case $(II)$: For a loop of radius $2r$,the length is $l_2 = 2\pi(2r) = 2l_1$. The resistance is $R_2 = \rho \frac{l_2}{A} = 2R_1$. The current is $I_2 = \frac{V}{R_2} = \frac{V}{2R_1} = \frac{I_1}{2}$.The new magnetic field is $B_2 = \frac{\mu_0 I_2}{2(2r)} = \frac{\mu_0 (I_1/2)}{4r} = \frac{1}{4} \left( \frac{\mu_0 I_1}{2r} \right) = \frac{B_1}{4}$.

Similar Questions

$A$ current of $1\,A$ is flowing through the sides of an equilateral triangle of side $4.5\times10^{-2}\,m$. The magnetic field at the centre of the triangle will be

$Weber/m^2$ is equal to

$A$ long thin hollow metallic cylinder of radius '$R$' carries a current '$i$'. The magnetic induction '$B$' at a distance '$r$' from the axis varies as shown in:

Find the magnitude of the magnetic field at point $p$ due to the semi-infinite wire shown below.

$A$ long straight wire carries an electric current of $2\,A$. The magnetic induction at a perpendicular distance of $5\,m$ from the wire is

Vedclass Test Series

Exam Paper Generator

Online Exam Module