A square wire 2.0 cm on each side is oriented so that a 5.00 mT magnetic field makes a 30.0º angle with the normal to the plane of the square.If a 1.0 mA current causes a 32.0 N · m torque on the wire,how many loops does the wire make? A)3.2 × 10¹⁰ B)3.2 × 10⁶ C)1.6 × 10¹⁰ D)4.5 × 10¹⁰

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Quizplus · Accepted Answer

The torque on a current loop is given by:$$	au = NIAB\sin	heta$$where:* N is the number of loops* I is the current* A is the area of the loop* B is the magnetic field strength* θ is the angle between the normal to the loop and the magnetic fieldIn this problem, we are given:* I = 1.0 mA = 0.001 A* A = (2.0 cm)^2 = 4.0 cm^2 = 4.0 x 10^-4 m^2* B = 5.00 mT = 0.005 T* θ = 30.0º* τ = 32.0 N · mWe can solve for N by rearranging the equation:$$N = \frac{	au}{IAB\sin	heta}$$Plugging in the values we know, we get:$$N = \frac{32.0 	ext{ N} \cdot 	ext{m}}{(0.001 	ext{ A})(4.0 	imes 10^{-4} 	ext{ m}^2)(0.005 	ext{ T})(\sin 30.0º)}$$$$N = 3.2 	imes 10^{10}$$Therefore, the correct answer is A.

A Square Wire 2