Consider a long straight wire of a circular c | vedclass

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

Question

(B) The maximum magnetic field $B_{\max}$ for a long straight wire occurs at its surface $(r = a)$: $B_{\max} = \frac{\mu_0 I}{2 \pi a}$. We are looki

Vedclass · Accepted Answer

$[a/2, 2a]$

Vedclass · Answer

(B) The maximum magnetic field $B_{\max}$ for a long straight wire occurs at its surface $(r = a)$: $B_{\max} = \frac{\mu_0 I}{2 \pi a}$. We are looking for the distances where the magnetic field is half of this value,i.e.,$B = \frac{B_{\max}}{2} = \frac{\mu_0 I}{4 \pi a}$. Inside the wire $(r Setting $B_{	ext{in}} = \frac{\mu_0 I}{4 \pi a}$,we get $\frac{\mu_0 I r}{2 \pi a^2} = \frac{\mu_0 I}{4 \pi a}$,which simplifies to $r = \frac{a}{2}$. Outside the wire $(r > a)$,the magnetic field is given by $B_{	ext{out}} = \frac{\mu_0 I}{2 \pi r}$. Setting $B_{	ext{out}} = \frac{\mu_0 I}{4 \pi a}$,we get $\frac{\mu_0 I}{2 \pi r} = \frac{\mu_0 I}{4 \pi a}$,which simplifies to $r = 2a$. Thus,the distances are $r = \frac{a}{2}$ (inside) and $r = 2a$ (outside).

Similar Questions

$PQ$ and $RS$ are long parallel conductors separated by a certain distance. $M$ is the midpoint between them (see the figure). The net magnetic field at $M$ is $B$. Now,the current $2 \text{ A}$ is switched off. The field at $M$ now becomes

Two circular coils $P$ and $Q$ are made from similar wire,but the radius of $Q$ is twice that of $P$. What should be the value of the potential difference across them so that the magnetic induction at their centres is the same?

In the figure,two parallel infinitely long current-carrying wires are shown. If the resultant magnetic field at point $A$ is zero,then determine the current $I$ (in $A$).

Vedclass Test Series

Exam Paper Generator

Online Exam Module