Question

A circular coil 14.0 cm in radius and composed of 145 tightly wound turns carries a current of 2.80 A in the counterclockwise direction, where the plane of the coil makes an angle of 15.0° with the y axis (see figure below). The coil is free to rotate about the z axis and is placed in a region with a uniform magnetic field given by B = 1.40 ſ T. B O = 15.0° (a) What is the magnitude of the magnetic torque on the coil? N·m (b) In what direction will the coil rotate? counterclockwise as seen from the +z axis clockwise as seen from the +z axis

Answer 1

Area of circular coil is

$A = \pi r^2 \\ A = 3.14*(0.14)^2 = 0.061544~m^2$

Angle between magnetic field and unit area vector is 90-15 = 75 degree

Torque $\tau = BINA~sin(\theta)$

             $\tau = 1.4*2.8*145*0.061544*~sin(75) \\ \tau = 33.79 ~N.m$

(b) counterclockwise