(b) A 2.0 m length of straight wire carries current of 20 A in a uniform magneti
ID: 1432566 • Letter: #
Question
(b) A 2.0 m length of straight wire carries current of 20 A in a uniform magnetic field of 50 mT whose direction is at an angle of 37^o from the direction of the current. (i) Find the force on the wire. (ii) If this wire is converted into a 6 turns-rectangular coil 4.3 cm wide and 5.8 cm long, carrying the same current in the same magnetic field, how much torque is required to hold the coil so that it makes an angle of 45^o with the magnetic field? (iii) How much work is required to rotate the coil 180^o?
Explanation / Answer
l =2 m , i =20 A, B =50 mT, theta =37 degrees
(i) F = ilBsin(theta) = 20*2*0.05*sin(37)
F = 1.2 N
(ii) A = lb = 4.3*5.8*10^-4 = 24.94*10^-4 m^2
magnetic moment M = iA = 20*24.94*10^-4 = 49.88*10^-3 A.m^2
Torque T =MBsin(theta) = 49.88*10^-4*0.05*sin(45)
Torque = 1.764*10^-4 N.m
(c) W = -MBcos(theta)
W = - 49.88*10^-4*0.05*cos(180)
W = 2.494*10^-4 J
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