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Introduction to Electrodynamics
Found in: Page 228
Introduction to Electrodynamics

Introduction to Electrodynamics

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

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Short Answer

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 μ0l81a-1b pointed outward.

(b) The magnetic field of a semi circular wire of radius R extending to infinity at each end and carrying current l μ0l4R1+2π pointed inward.

See the step by step solution

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

B=μ0l2R .....(1)

Here, μ0 is the permeability of free space.

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

B=μ0l2πR ....(2)

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=14×μ0l2a =μ0l8a

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=14×μ0l2b =μ0l8b

This field is pointed inward according to right hand rule.

Thus, the net field at P is

B=μ0l8(1a-1b)

The field is pointed outward.

The net field at P is μ0l8(1a-1b) 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=μ0l2πR

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

B=12×μ0l2R =μ0l4R

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

B=μ0l2πR+μ0l4R =μ0l4R1+2π

Thus, the net field at P is μ0l4R1+2π pointed inwards.

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