There are 124 turns in a flat coil of wire that is immersed in a uniform magneti
ID: 1444419 • Letter: T
Question
There are 124 turns in a flat coil of wire that is immersed in a uniform magnetic field and the coil carries a current of 2.09 mA. If the angle between the normal to the loop and the magnetic field is 40.3degree, the area of the coil is 8.01 cm^2 and the torque acting on the coil is 1.45 times 10^-5 N-m, find the magnetic field strength. How are the number of turns, the area of the loop, the angle between the magnetic dipole moment and the field, the current in the loop, and the magnetic field strength related to the torque?Explanation / Answer
T = n*I*A*B*sin thetha
1.45*10^-5 = 124*(2.09*10^-3)*(8.01*10^-4)*B*sin 40.3
B =0.108 T
Answer: 0.108 T
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