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Q9P

Expert-verifiedFound in: Page 228

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.

(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 .

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

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

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

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

This field is pointed outward according to right hand rule.

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

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

This field is pointed inward according to right hand rule.

Thus, the net field at P is

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

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.

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

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

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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