A wire of length l is bent into a circular lo | 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 loop of radius $R$ with $N$ turns carrying current $I$ is given by $B = \frac{\mu_0 N I}{2 R}$.For

Vedclass · Accepted Answer

$4 B$

Vedclass · Answer

(C) The magnetic field at the centre of a circular loop of radius $R$ with $N$ turns carrying current $I$ is given by $B = \frac{\mu_0 N I}{2 R}$.For a single loop $(N=1)$,$B = \frac{\mu_0 I}{2 R}$. Since the length of the wire $l = 2 \pi R$,we have $R = \frac{l}{2 \pi}$.Thus,$B = \frac{\mu_0 I}{2 (l / 2 \pi)} = \frac{\mu_0 I \pi}{l}$.When the same wire of length $l$ is bent into a double loop $(N=2)$,the new radius $R'$ is given by $l = 2 (2 \pi R')$,so $R' = \frac{l}{4 \pi} = \frac{R}{2}$.The new magnetic field $B'$ at the centre is $B' = \frac{\mu_0 N' I}{2 R'} = \frac{\mu_0 (2) I}{2 (R / 2)} = \frac{2 \mu_0 I}{R} = 4 \left( \frac{\mu_0 I}{2 R} \right) = 4 B$.

Similar Questions

Consider three observations $a, b$ and $c$ such that $b = a + c$. If the standard deviation of $a + 2, b + 2, c + 2$ is $d$,then which of the following is true?

The portion $AB$ of the indicator diagram representing the state of matter denotes:

If $f(x) = \cot^{-1} \left( \frac{x^x - x^{-x}}{2} \right)$,then $f'(1)$ is equal to

In an unbiased $p-n$ junction,holes diffuse from the $p-$region to $n-$region because

Vedclass Test Series

Exam Paper Generator

Online Exam Module