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

Question

Toppr 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

$n \times 2 π r = 2 π R$

A long wire carrying a steady current is bent into a circular loop of one turn. The magnetic filed at the centre of the loop is B. It is them bent into circular coil of n turns. The magnetic field at the centre of this cold of n turns will be.2n2Bn2BnB2n=B

The circumference of the first loop is the length of the wire=2πRThe magnetic field at the center of this wire=B=μ0I2RThe same wire is bent into n circular coils.Thus new radius is found out byn×2πr=2πR

A long wire carrying a steady current is bent into a circular loop of one turn. The magnetic filed at the centre of the loop is B. It is them bent into circular coil of n turns. The magnetic field at the centre of this cold of n turns will be.
$2 n^{2} B$
$n^{2} B$
$n B$
$2 n = B$

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

$n \times 2 π r = 2 π R$