A rectangular coil of wire 22.0 cm by 35.0 cm and carrying a current of 1.40 A i
ID: 2006597 • Letter: A
Question
A rectangular coil of wire 22.0 cm by 35.0 cm and carrying a current of 1.40 A is oriented with the plane of its loop perpendicular to a uniform 1.50 T magnetic field as shown in the figure. = 2f
Calculate the torque that the magnetic field exerts on the coil.
The coil is now rotated through a 30.0 degree angle about the axis shown, the left side coming out of the plane and the right side going into the plane. Calculate the torque that the magnetic field exerts on the coil.
Now, the rectangular coil of wire is held fixed in its original position in the figure above with the magnetic field perpendicular to the plane of the rectangular loop. The magnetic field is increased from 1.50 T to 3.10 T in a time of 0.6 s. Calculate the magnitude of the induced emf.
Now, the magnetic field is held constant at 3.10 T and the rectangular loop is rotated about the axis at 30 Hz. Calculate the magnitude of the maximum induced emf.
Explanation / Answer
The area of the loop A = 22cm * 35cm = 770cm^2 or 0.077m^2 The current in the coil I = 1.40A the magnetic field B = 1.50T (a) The torque = NIAB sin = (1)(1.4)(0.077) (1.5)sin90 = 0.162 N.m (b) When the coil rotates 30 deg about the axis the torque = NIAB sin = (1)(1.4)(0.077) (1.5)sin 30 = 0.081 N.m = (1)(1.4)(0.077) (1.5)sin 30 = 0.081 N.m (c) When the field changes from 1.5T to 3.1T in time t = 0.6s the induced emf = - d/dt = - A dB/dt cos0 = - (0.077)(1.5-3.1/0.6) = 0.205 V (d) When the field B = 3.1T and frequency f = 30Hz therefore the maximum emf = NAB = (1)(0.077)(3.1)(2*30) = 14.5 VRelated Questions
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