Question

4..+ .11%92 18:27 Side view conducting coil consisting of n turns of wire is placed ina uniform magnetic field of magnitude B 1.61T Top view oriented in the positive y direction as shown in the figure The radius of the coil is R-5.43 cm, and the angle between the magnetic field vector and the unit normal vector to the coil is θ-60.0°. The current through the coil is i-5.55 A. 2R 60° a) Calculate the number of turns, n, the coil must have for the torque on the loop to be 2.61 N m 36.41 Computer's answer now shown above. Tries 0/99 b) If the radius of the loop is decreased to R 1.63 cm, what should the number of turns, N, be for the torque to remain unchanged? Assume that i, B, and 0 stay the same

Answer 1

From given data, Magnitude = B = 1.01 T Radius = R = 5.43 cm = 5.43 times 10^-2 m theta = 60 degrees 0 Current through the coil = I = 5.55A Area = A = Pi R^2 (a) Magnetic Moment = mu = IAN where N = No. of turns Z = 2.61 N middot m Torque = T = mu^vector times B^vector = mu B sin theta T = NIAB sin theta N = T/IAB sin theta = 2 times 2.61/squareroot 3 (5.55) Pi (5.43 times 10^-2)^2 times 1.61 N almostequalto 36.41 Turns (b) If R = 1.63 Cm = 1.63 times 10^-2 m T = NIAB sin theta N = T/IAB sin theta N = 2 times 2.61/squareroot 3 times (5.55)Pi times (1.63 times 10^-2)^2 times 1.61 N almostequalto 404 turns

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