The magnetic field at the centre of a single- | vedclass

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

Question

(C) Let the total length of the wire be $L$. For a single-turn coil $(n_1 = 1)$,the circumference is $L = 2\pi R_1$,so $R_1 = \frac{L}{2\pi}$. The mag

Vedclass · Accepted Answer

$4B$

Vedclass · Answer

(C) Let the total length of the wire be $L$. For a single-turn coil $(n_1 = 1)$,the circumference is $L = 2\pi R_1$,so $R_1 = \frac{L}{2\pi}$. The magnetic field at the centre is $B_1 = \frac{\mu_0 I}{2R_1} = \frac{\mu_0 I}{2(L/2\pi)} = \frac{\mu_0 I \pi}{L} = B$. For a two-turn coil $(n_2 = 2)$,the total length $L = n_2(2\pi R_2) = 2(2\pi R_2) = 4\pi R_2$. Thus,the new radius is $R_2 = \frac{L}{4\pi} = \frac{R_1}{2}$. The magnetic field at the centre for $n$ turns is $B_n = \frac{n \mu_0 I}{2R_n}$. For $n=2$,$B_2 = \frac{2 \mu_0 I}{2R_2} = \frac{\mu_0 I}{R_2}$. Substituting $R_2 = \frac{R_1}{2}$,we get $B_2 = \frac{\mu_0 I}{R_1/2} = \frac{2 \mu_0 I}{R_1}$. Since $B = \frac{\mu_0 I}{2R_1}$,we have $\frac{\mu_0 I}{R_1} = 2B$. Therefore,$B_2 = 2(2B) = 4B$.

The magnetic field at the centre of a single-turn circular coil of a given length of wire carrying a current $I$ is $B$. If the same wire is bent into a circular coil of two turns and the same current $I$ is passed through it,the new magnetic field at the centre will be:

Similar Questions

The magnetic dipole moment of a rectangular current-carrying loop is:

$A$ rod of length $l$ having uniformly distributed charge $Q$ is rotated about one end with constant frequency $f$. Its magnetic moment is:

Show the magnetic field lines for a current-carrying loop. Write the rule for finding the direction of a magnetic field due to a circular current loop.

$A$ wire of length $L$ metre carrying a current of $I$ ampere is bent in the form of a circle. Its magnetic moment is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module