A long wire carrying a steady current is bent into a circular loop of one turn. The magnetic field at the centre of the loop is B. It is then bent into a circular coil of n turns. The magnetic field at the centre of this coil of n turns will be

Question

A long wire carrying a steady current is bent into a circular loop of one turn. The magnetic field at the centre of the loop is B. It is then bent into a circular coil of n turns. The magnetic field at the centre of this coil of n turns will be (A) nB (B) n2 B (C) 2nB (D) 2n2 B

Tardigrade Admin · Accepted Answer

The circumference of the first loop is the length of the wire

= 2 π R

The magnetic field at the center of this wire

= B = \frac{μ _{0} I}{2 R}

The same wire is bent into

n

circular coils.
Thus new radius is found out by

n \times 2 π r = 2 π R

⟹ r = \frac{R}{n}

Thus magnetic field at the center of these loops

= n \frac{μ _{0} I}{2 r}

= \frac{n ^{2} μ _{0} I}{2 R} = n^{2} B

Q. A long wire carrying a steady current is bent into a circular loop of one turn. The magnetic field at the centre of the loop is $B$ . It is then bent into a circular coil of $n$ turns. The magnetic field at the centre of this coil of $n$ turns will be

$n B$

$n^{2} B$

$2 n B$

$2 n^{2} B$

Q. A long wire carrying a steady current is bent into a circular loop of one turn. The magnetic field at the centre of the loop is B. It is then bent into a circular coil of n turns. The magnetic field at the centre of this coil of n turns will be

nB

n2B

2nB

2n2B

Q. A long wire carrying a steady current is bent into a circular loop of one turn. The magnetic field at the centre of the loop is $B$ . It is then bent into a circular coil of $n$ turns. The magnetic field at the centre of this coil of $n$ turns will be

$n B$

$n^{2} B$

$2 n B$

$2 n^{2} B$