A52.5-turn circular coil of radius 5.35 cm can be oriented in any direction in a uniform magnetic field having a magnitude of 0.455 t. if the coil carries a current of 25.3 ma, find the magnitude of the maximum possible torque exerted on the coil.

5.43 x 10^-3 Nm

Explanation:

N = 52.5, radius, r = 5.35 cm = 0.0535 m, B = 0.455 T, I = 25.3 mA = 0.0253 A

Torque = N I A B Sin theta

Here, theta = 90 degree

Torque = 52.5 x 0.0253 x 3.14 x 0.0535 x 0.0535 x 0.455

Torque = 5.43 x 10^-3 Nm

b

step-by-step explanation: a third of 24 is 8, 8 times 3 is 24 so 8 - 2 would be 6

answer: 2×10^-2kg

explanation: density= mass per volume

vol=mass/density

converting to s.i units we have;

3÷1000kg = 0.003

1.5÷100m = 0.15

0.003/0.15 = 0.02kg