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Chapter 29: Q. 17 (page 831)

A 100Acurrent circulates around a 2.0mmdiameter superconducting ring.

a. What is the ring’s magnetic dipole moment?

b. What is the on-axis magnetic field strength 5.0cmfrom the ring?

Short Answer

Expert verified

a. The ring magnetic moment is 3.14×10-4A.m2.

b. the magnetic field strength is0.5μT.

Step by step solution

01

Part (a) Step 1: Given information

We have given,

current = 100A

Diameter = 2mm

We have to find the magnetic dipole moment.

02

Simplify

The magnetic dipole moment of ring is

μ=AIμ=πr2Iμ=3.14×10-6×100μ=3.14×10-4A.m2

03

Part (b) Step 1: Given information

We have given,

Distance = 5cm

We have to find the magnetic field strength.

04

Simplify

Magnetic field due to dipole is

B=μ0μ→2πR3B=(4π×10-7)(3.14×10-4)2π(5×10-2)3B=0.5μT

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