$A$ long wire carrying a steady current is bent into a circle of single turn. The magnetic field at the centre of the coil is $B$. If it is bent into a circular loop of radius $r_1$ having $n$ turns,the magnetic field at the centre of the coil for the same current is:

  • A
    $B/n^2$
  • B
    $B/n$
  • C
    $n^2 B$
  • D
    $n B$

Explore More

Similar Questions

The direction of magnetic lines of force produced by passing a direct current in a conductor is given by

$A$ steady current $I$ flows through a wire loop $PQR$ having the shape of a right-angled triangle with $PQ = 3x$, $PR = 4x$, and $QR = 5x$. If the magnitude of the magnetic field at $P$ due to this loop is $k \left( \frac{\mu_0 I}{48 \pi x} \right)$, find the value of $k$.

An equilateral triangle of side '$a$' carries a current '$i$'. What is the magnetic field at point '$P$' (a vertex of the triangle)?

$A$ $10 \ A$ current is passing through a very long wire of radius $5 \ cm$. The magnetic field at a distance of $2 \ cm$ inside from its curved surface is . . . . . . $\times 10^{-5} \ T$.

$A$ circular arc of radius $r$ carrying current $I$ subtends an angle $\frac{\pi}{16}$ at its centre. The radius of the metal wire is uniform. The magnetic induction at the centre of the circular arc is [where $\mu_0$ is the permeability of free space].

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo