A 20-cm-diameter loop of wire has a resistance of 150 . Itis initially in a unif

Question

A 20-cm-diameter loop of wire has a resistance of 150 . Itis initially in a uniform 0.40-T magnetic fi eld

with its normal at an angle of 30° with respect to the fi eld.In a time interval of 0.1 s, the loop is turned

until its normal aligns with the B-fi eld. Then, in an additionaltime interval of 0.18 s, the loop is completely

withdrawn (without turning) from the fi eld. Calculate the totalelectrical energy dissipated during this entire

two-step process.

I dont understand this question at all so please explain.

A 20-cm-diameter loop of wire has a resistance of 150 . Itis initially in a uniform 0.40-T magnetic fi eld

with its normal at an angle of 30° with respect to the fi eld.In a time interval of 0.1 s, the loop is turned

until its normal aligns with the B-fi eld. Then, in an additionaltime interval of 0.18 s, the loop is completely

withdrawn (without turning) from the fi eld. Calculate the totalelectrical energy dissipated during this entire

two-step process.

I dont understand this question at all so please explain.

Explanation / Answer

the force acting on the wire is F = -I * l * B or m * g = -I * l * B or I = (m * g/l * B) --------------(1) m = * V = * (4/3) * (d/2)^3 = (/6) * * d^3 is the density of copper wire and d is the diameter ofwire d = 20 cm = 20 * 10^-2 m g = 9.8 m/s^2 l is the length of the copper wire B is the magnetic field and is equal to 0.40 T The total electrical energy dissipated during the entiretwo-step process is P = I^2 * R the value of I is obtained from equation (1) and R = 150

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