Question

Calculate the magnetic field strength needed on a 200-turn square loop 20.0 cm on a side to create a maximum torque of $300\textrm{ N}\cdot\textrm{m}$ if the loop is carrying 25.0 A.

Final Answer

$1.50\textrm{ T}$

### Solution video

# OpenStax College Physics for AP® Courses, Chapter 22, Problem 44 (Problems & Exercises)

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Video Transcript

This is College Physics Answers with Shaun Dychko. We are going to calculate the magnetic field strength needed to have the maximum torque on this square loop given that the loop has a current of 25.0 amps flowing through it, a side length of 20.0 centimeters which is 20.0 times 10 to the minus 2 meters and 200 turns of wire and the maximum torque is 300 newton meters. So the maximum torque is the number of turns multiplied by the current multiplied by the area of the loop times the magnetic field strength so we can solve for

*B*by dividing both sides by*NIA*. So*B*is the maximum torque divided by number of turns times the current times the area of the loop. Now since this loop is a square that means its area is its side length multiplied by itself so side length squared so we can substitute that in for*A*. So the magnetic field strength then is 300 newtons meters divided by 200 turns times 25.0 amps times 20.0 times 10 to the minus 2 meters squared, which gives 1.50 tesla.