A 25-turn square coil with a mass of 0.320 kg and a side of 0.210 m is hinged along a horizontal side,carries a 6.00-A current and is placed in a vertical magnetic field pointing down and having a magnitude of 4.00 mT.Determine the angle that the plane of the coil makes with the vertical when the coil is in equilibrium.Use g = 9.81 m/s².

Question

Quizplus · Accepted Answer

The torque due to the magnetic field is:$$	au = NIAB\sin	heta$$where:* N is the number of turns* I is the current* A is the area of the coil* B is the magnetic field strength* 	heta is the angle between the plane of the coil and the magnetic fieldThe torque due to gravity is:$$	au = mgl\sin	heta$$where:* m is the mass of the coil* g is the acceleration due to gravity* l is the length of the side of the coilWhen the coil is in equilibrium, the two torques are equal:$$NIAB\sin	heta = mgl\sin	heta$$Solving for 	heta, we get:$$	heta = 	an^{-1}\left(\frac{mgl}{NIAB}ight)$$Plugging in the given values, we get:$$	heta = 	an^{-1}\left(\frac{(0.320	ext{ kg})(9.81	ext{ m/s}^2)(0.210	ext{ m})}{(25)(6.00	ext{ A})(0.210	ext{ m})^2(4.00	imes10^{-3}	ext{ T})}ight)$$$$	heta = 4.6^\circ$$

A 25-Turn Square Coil with a Mass of 0 ∘\circ∘ B)2

A 25-Turn Square Coil with a Mass of 0 $\circ$
B)2