Acircular coil of radius 8.0 cm and 20 turns is rotated about itsvertical diameter with an angular speed of 50 rad s−1in a uniform horizontal magnetic field of magnitude 3.0×10−2T. Obtain the maximum and average emf induced in the coil. If thecoil forms a closed loop of resistance 10Ω,calculate the maximum value of current in the coil. Calculate theaverage power loss due to Joule heating. Where does this power comefrom?
Acircular coil of radius 8.0 cm and 20 turns is rotated about itsvertical diameter with an angular speed of 50 rad s−1in a uniform horizontal magnetic field of magnitude 3.0×10−2T. Obtain the maximum and average emf induced in the coil. If thecoil forms a closed loop of resistance 10Ω,calculate the maximum value of current in the coil. Calculate theaverage power loss due to Joule heating. Where does this power comefrom?
Max induced emf =0.603 V
Average induced emf= 0 V
Max current in the coil= 0.0603 A
Average power loss =0.018 W
(Power comes from theexternal rotor)
Radius of the circularcoil, r = 8 cm = 0.08 m
Area of the coil, A= πr2 = π× (0.08)2 m2
Number of turns on thecoil, N = 20
Angular speed, ω= 50 rad/s
Magnetic fieldstrength, B = 3 ×10−2 T
Resistance of the loop,R = 10 Ω
Maximum induced emfis given as:
e = NωAB
= 20 ×50 × π× (0.08)2 ×3 × 10−2
= 0.603 V
The maximum emfinduced in the coil is 0.603 V.
Over a full cycle, theaverage emf induced in the coil is zero.
Maximum current isgiven as:
Average power loss dueto joule heating:
The current induced inthe coil produces a torque opposing the rotation of the coil. Therotor is an external agent. It must supply a torque to counter thistorque in order to keep the coil rotating uniformly. Hence,dissipated power comes from the external rotor.
Max induced emf =0.603 V
Average induced emf= 0 V
Max current in the coil= 0.0603 A
Average power loss =0.018 W
(Power comes from theexternal rotor)
Radius of the circularcoil, r = 8 cm = 0.08 m
Area of the coil, A= πr2 = π× (0.08)2 m2
Number of turns on thecoil, N = 20
Angular speed, ω= 50 rad/s
Magnetic fieldstrength, B = 3 ×10−2 T
Resistance of the loop,R = 10 Ω
Maximum induced emfis given as:
e = NωAB
= 20 ×50 × π× (0.08)2 ×3 × 10−2
= 0.603 V
The maximum emfinduced in the coil is 0.603 V.
Over a full cycle, theaverage emf induced in the coil is zero.
Maximum current isgiven as:
Average power loss dueto joule heating:
The current induced inthe coil produces a torque opposing the rotation of the coil. Therotor is an external agent. It must supply a torque to counter thistorque in order to keep the coil rotating uniformly. Hence,dissipated power comes from the external rotor.
