Problem 20.51 A single square loop of wire 20.5 cm on a side is placed with its

Question

Problem 20.51

A single square loop of wire 20.5 cm on a side is placed with its face parallel to the magnetic field as in (Figure 1) . When I = 5.20 A flows in the coil, the torque on it is 0.325 mN .

Part A

What is the magnetic field strength?

Express your answer using three significant figures.

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Figure 1 of 1

Problem 20.51

A single square loop of wire 20.5 cm on a side is placed with its face parallel to the magnetic field as in (Figure 1) . When I = 5.20 A flows in the coil, the torque on it is 0.325 mN .

Part A

What is the magnetic field strength?

Express your answer using three significant figures.

SubmitHintsMy AnswersGive UpReview Part

Provide FeedbackContinue

B = T

Figure 1 of 1

Explanation / Answer

torque = N*i*A*B*sin theta

The face of the loop of wire is parallel to the magnetic field, the angle between the perpendicular to the loop and the magnetic field is 90 deg

sin 90 deg = 1

B = torque/NiA

A = side^2 = 0.205*0.205

B = 0.325/(1*5.20*0.205*0.205) = 1.487 T

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