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Chapter 30: Q66P (page 900)

A circular loop of wire 50 mmin radius carries a current of 100 A. (a) Find the magnetic field strength. (b) Find the energy density at the center of the loop

Short Answer

Expert verified

a)B=1.3脳10-3Tb)uB=0.63Jm3

Step by step solution

01

Given

i) Radius of the circular loop R=50mm=50脳10-3m

ii) Current in the circular loopi=100A

02

Understanding the concept

We can use the formula of magnetic field at the center of the current loop to find it. Using this magnetic field into the formula of the energy density at that point, we can find the energy density at the center of the loop.

Formula:

B=渭0i/2RuB=B2/2渭0

03

(a) Calculate the magnetic field strength

The magnetic field at the center of loop is given by

B=渭0i2RB=4脳10-7Hm100A250脳10-3mB=1.3脳10-3T

Therefore, the magnetic field strength is 1.3脳10-3T.

04

(b) Calculate the energy density at the center of the loop

The energy density at the centre of the loop is given by

uB=B22渭0uB=1.3脳10-3T224蟺脳10-7HmuB=0.63Jm3

Therefore, energy density at the center of the loop is role="math" localid="1661424939827" 0.63J/m3.

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Most popular questions from this chapter

In Figure, 蔚=100V,R1=10.0惟,R2=20.0惟,R3=30.0惟,andL=2.00H.Immediately after switch S is closed, (a) what is i1? (b) what is i2? (Let currents in the indicated directions have positive values and currents in the opposite directions have negative values.) A long time later, (c) what is i1? (d) what is i2? The switch is then reopened. Just then, (e) what is i1? (f) what is i2? A long time later, (g) what is i1? (h) what is i2?

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