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A current of 19.5 mA is maintained in a single circular loop with a circumference of 2.30 m. A magnetic field of 0.710 T is directed paralle
Question
A current of 19.5 mA is maintained in a single circular loop with a circumference of 2.30 m. A magnetic field of 0.710 T is directed parallel to the plane of the loop. What is the magnitude of the torque exerted by the magnetic field on the loop?
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Physics
3 years
2021-08-17T18:29:14+00:00
2021-08-17T18:29:14+00:00 1 Answers
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Answers ( )
Answer:
0.0058 Nm
Explanation:
Parameters given:
Number of turns, N = 1
Current, I = 19.5 mA = 0.0195 A
Circumference = 2.3 m
Magnetic field, B = 0.710 T
Angle, θ = 90°
Magnetic torque is given as:
τ = N * I * A * B * sinθ
Where A is area
Circumference is given as:
C = 2 * pi * r
=> 2.3 = 2 * pi * r
=> r = 2.3/(2 * 3.142)
r = 0.366 m
Area, A can now be found:
A = pi * r² = 3.142 * 0.366²
A = 0.42 m²
Therefore, torque is:
τ = 1 * 0.0195 * 0.42 * 0.71 * sin90
τ = 0.0058 Nm