A 10.0-m-long copper wire with cross-sectional area 1.25 × 10^-4 m² is bent into a square loop,and then connected to a 0.25-V battery.The loop is then placed in a uniform magnetic field of magnitude 0.19 T.What is the maximum torque that can act on this loop? The resistivity of copper is 1.70 × 10^-8Ω·m.

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Quizplus · Accepted Answer

The maximum torque occurs when the magnetic field is perpendicular to the plane of the loop. The torque is given by:$$	au = IAB\sin	heta$$where I is the current, A is the area of the loop, B is the magnetic field, and θ is the angle between the magnetic field and the normal to the plane of the loop.The current is given by:$$I = \frac{V}{R}$$where V is the voltage and R is the resistance.The resistance is given by:$$R = \frac{ho L}{A}$$where ρ is the resistivity, L is the length of the wire, and A is the cross-sectional area of the wire.Substituting these equations into the equation for torque, we get:$$	au = \frac{VAB\sin	heta}{ho L}$$Plugging in the given values, we get:$$	au = \frac{(0.25 V)(1.25 	imes 10^{-4} m^2)(0.19 T)\sin 90^\circ}{1.70 	imes 10^{-8} \Omega \cdot m (10.0 m)}$$$$	au = 2.2 	imes 10^{-2} N \cdot m$$

A 100-M-Long Copper Wire with Cross-Sectional Area 1