Question

Question: Find the magnetic field at point Pfor each of the steady current configurations shown in Fig. 5.23.

Short answer

(a) The magnetic field at the center of a quarter circular ring of inner radius a and outer radius and carrying current l is $\frac{{\mu }_{0}l}{8}\left(\frac{1}{a}-\frac{1}{b}\right)$pointed outward.

(b) The magnetic field of a semi circular wire of radius R extending to infinity at each end and carrying current$l\frac{{\mu }_{0}l}{4R}\left(1+\frac{2}{\mathrm{\pi }}\right)$ pointed inward.

Step-by-step solution

Given data

(a) A quarter circular ring of inner radius a and outer radius b and carrying current l .

(b) A semi circular wire of radius R extending to infinity at each end and carrying current l .

Magnetic field from a circle and infinite straight wire

The field due to a circle of radius and carrying current at the center is

$\mathbf{B}\mathbf{=}\frac{{\mathbf{\mu }}_{\mathbf{0}}\mathbf{l}}{\mathbf{2}\mathbf{R}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\left(1\right)$

Here, ${\mu }_0$is the permeability of free space.

The field due to an infinite straight wire carrying current l at a distance R from it is

$\mathbf{B}\mathbf{=}\frac{{\mathbf{\mu }}_{\mathbf{0}}\mathbf{l}}{\mathbf{2}\mathbf{\pi R}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\left(2\right)$

Magnetic field from figure (a)

In the first figure, the straight sections produce no field at P because their extended sections pass through it.

From equation (1), the field from the inner ring is

$$\begin{gathered} B=\frac{1}{4}\times \frac{{\mu }_{0}l}{2a} \\ =\frac{{\mu }_{0}l}{8a} \end{gathered}$$

This field is pointed outward according to right hand rule.

From equation (1), the field from the outer ring is

role="math" localid="1657774429401" $$\begin{gathered} B=\frac{1}{4}\times \frac{{\mu }_{0}l}{2b} \\ =\frac{{\mu }_{0}l}{8b} \end{gathered}$$

This field is pointed inward according to right hand rule.

Thus, the net field at P is

$\mathit{B}\mathbf{=}\frac{{\mathbf{\mu }}_{\mathbf{0}}\mathbf{l}}{\mathbf{8}}\left(\frac{1}{a}-\frac{1}{b}\right)$

The field is pointed outward.

The net field at P is $\frac{{\mathit{\mu }}_{\mathbf{0}}\mathit{l}}{\mathbf{8}}\mathbf{(}\frac{\mathbf{1}}{\mathit{a}}\mathbf{-}\frac{\mathbf{1}}{\mathit{b}}\mathbf{)}$pointed outward.

Magnetic field from figure (b)

The two half infinite sections at the top and bottom of the second figure form an infinite wire with field from equation (2) at P

$B=\frac{{\mu }_{0}l}{2\pi R}$

The field from the semi circular section from equation (1) at P is

$$\begin{gathered} B=\frac{1}{2}\times \frac{{\mu }_{0}l}{2R} \\ =\frac{{\mu }_{0}l}{4R} \end{gathered}$$

Both of these fields are pointed inwards. Thus the net field at P is

$$\begin{gathered} B=\frac{{\mu }_{0}l}{2\pi R}+\frac{{\mu }_{0}l}{4R} \\ =\frac{{\mu }_{0}l}{4R}\left(1+\frac{2}{\pi }\right) \end{gathered}$$

Thus, the net field at P is $\frac{{\mu }_{0}l}{4R}\left(1+\frac{2}{\pi }\right)$pointed inwards.