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

The magnetic field is needed to be 1.5 $\mathrm{T}$

Physics 102 Electricity and Magnetism

Chapter 22

Magnetism

Magnetic Field and Magnetic Forces

Sources of Magnetic field

Electromagnetic Induction

Inductance

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here. We know that the Twerk the equation for the torque is gonna be equaling the number of turns in the loop multiplied by eye, the current multiplied by a the cross sectional area of one of of essentially each turn in the loop multiplied by be the magnitude of the magnetic field times sign of Fada. And so here this would simply be equaling. And I Here we have the length of the side of squared that would be substituted in for the area times be sign of Fada. And so we can then safe solving for the magnitude of the magnetic field. This would be equaling the torque divided by N i X squared sine of theater and we can solve. And so this would be the torque of 300 Newton meters, divided by the number of turns in the loop 200 times the current of 25.0 and peers times X squared, which in this case, 20 centimeters so 200.20 meters quantity squared, multiplied by sine of 90 degrees. And we find that be the magnitude of the magnetic field is equaling 1.50 Tesla's. This would be our final answer. That is the end of the solution. Thank you for Walter

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