A part of a long wire carrying a current i is bent into a circle of radius r as shown in figure. The net magnetic field at the centre O of the circular loop is

Question

A part of a long wire carrying a current i is bent into a circle of radius r as shown in figure. The net magnetic field at the centre O of the circular loop is (A)  (μ 0 i/4r) (B)  (μ 0 i/2r) (C)  (μ 0 i/2π r)(π +1)  (D)  (μ 0 i/2π r)(π -1)

Tardigrade Admin · Accepted Answer

The magnitude of the magnetic field at point O due to straight
part of wire is

B_{1} = \frac{μ _{0} i}{2 π r}

B_{1}

is perpendicular to the plane of the page, directed upwards
(right hand palm rule I).
The field at the centre O due to the current loop of radius r is

B_{2} = \frac{μ _{0} i}{2 r}

B_{2}

is also perpendicular to the page, directed upward (right
hand screw rule).

∴

Resultant field at O is

B_{1} + B_{2} = \frac{μ _{0} i}{2 r} (\frac{1}{π} + 1)

= \frac{μ _{0} i}{2 π r} (π + 1)

Q. A part of a long wire carrying a current i is bent into a
circle of radius r as shown in figure. The net magnetic
field at the centre O of the circular loop is

$\frac{μ _{0} i}{4 r}$

$\frac{μ _{0} i}{2 r}$

$\frac{μ _{0} i}{2 π r} (π + 1)$

$\frac{μ _{0} i}{2 π r} (π - 1)$

Q. A part of a long wire carrying a current i is bent into a circle of radius r as shown in figure. The net magnetic field at the centre O of the circular loop is

4rμ0​i​

2rμ0​i​

2πrμ0​i​(π+1)

2πrμ0​i​(π−1)

Q. A part of a long wire carrying a current i is bent into a
circle of radius r as shown in figure. The net magnetic
field at the centre O of the circular loop is

$\frac{μ _{0} i}{4 r}$

$\frac{μ _{0} i}{2 r}$

$\frac{μ _{0} i}{2 π r} (π + 1)$

$\frac{μ _{0} i}{2 π r} (π - 1)$