Question

A long thin current (I) carrying wire is bent in the form of a pin as shown in the figure. The magnetic field at point ‘O’ is: 

(μ₀ = permeability of free space)

Options

• `(mu_0 I)/(4 r) (1 - 2/pi)`

• `(mu_0 I)/(2 r) (1 + 2/pi)`

• `(mu_0 I)/(2 r) (1 - 2/pi)`

• `(mu_0 I)/(4 r) (1 + 2/pi)`

Answer 1

`bb((mu_0 I)/(4 r) (1 + 2/pi))`

Explanation:

Magnetic field due to the semi-infinite long wire is:

B = `(mu_0 I)/(4 pi r)`

There are two semi-infinite wires, and the magnetic field due to both of them is directed in the same direction.

∴ B = `(2 mu_0 I)/(4 pi r)`    ...(i)

Magnetic field due to a semi-circular wire at its centre.

B₂ = `(mu_0 I)/(4 pi r) xx theta`

= `(mu_0 I)/(4 pi r) xx pi`

= `(mu_0 I)/(4 r)`   ...(ii)

Net magnetic field at point O will be,

B_net = B₁ + B₂

= `(mu_0 I)/(4 r) (1 + 2/pi)`