A wire of length L carrying a current I is bent into a circle. The magnitude of the magnetic field at the centre of the circle is

Question

A wire of length L carrying a current I is bent into a circle. The magnitude of the magnetic field at the centre of the circle is (A) (π μ 0 I/L) (B) (μ 0 I/2 L) (C) (2 π μ0 I/L) (D) (μ 0 I/2 π L)

Tardigrade Admin · Accepted Answer

Length L= circumference of circle =2π R ⇒ R=(L/2 π ) ∴ Magnetic field at the centre of circular current loop is B=(μ 0 I/2 R)=(μ 0 I/2 L)× 2π B=(μ 0 π I/L)

Q. A wire of length $L$ carrying a current $I$ is bent into a circle. The magnitude of the magnetic field at the centre of the circle is

$\frac{π μ _{0} I}{L}$

$\frac{μ _{0} I}{2 L}$

$\frac{2 π μ _{0} I}{L}$

$\frac{μ _{0} I}{2 π L}$

Length $L =$ circumference of circle $= 2 π R$
$\Rightarrow R = \frac{L}{2 π}$
$∴$ Magnetic field at the centre of circular current loop is
$B = \frac{μ _{0} I}{2 R} = \frac{μ _{0} I}{2 L} \times 2 π$
$B = \frac{μ _{0} π I}{L}$

Q. A wire of length L carrying a current I is bent into a circle. The magnitude of the magnetic field at the centre of the circle is

Lπμ0​I​

2Lμ0​I​

L2πμ0​I​

2πLμ0​I​