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### Introduction to Electrodynamics

Book edition 4th edition
Author(s) David J. Griffiths
Pages 613 pages
ISBN 9780321856562

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

(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.

See the step by step solution

## Step 1: 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 .

## Step 2: 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

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

## Step 3: 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

$B=\frac{1}{4}×\frac{{\mu }_{0}l}{2a}\phantom{\rule{0ex}{0ex}}=\frac{{\mu }_{0}l}{8a}$

This field is pointed outward according to right hand rule.

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

role="math" localid="1657774429401" $B=\frac{1}{4}×\frac{{\mu }_{0}l}{2b}\phantom{\rule{0ex}{0ex}}=\frac{{\mu }_{0}l}{8b}$

This field is pointed inward according to right hand rule.

Thus, the net field at P is

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

The field is pointed outward.

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

## Step 4: 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

$B=\frac{1}{2}×\frac{{\mu }_{0}l}{2R}\phantom{\rule{0ex}{0ex}}=\frac{{\mu }_{0}l}{4R}$

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

$B=\frac{{\mu }_{0}l}{2\pi R}+\frac{{\mu }_{0}l}{4R}\phantom{\rule{0ex}{0ex}}=\frac{{\mu }_{0}l}{4R}\left(1+\frac{2}{\pi }\right)$

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

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