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Constant Magnetic Field, Magnetics

Problem 3.255

A square frame carrying a current I=0.90 AI=0.90 \mathrm{~A} is located in the same plane as a long straight wire carrying a current I0=I_{0}= =5.0=5.0 A. The frame side has a length a=8.0 cm.a=8.0 \mathrm{~cm} . The axis of the frame passing through the midpoints of opposite sides is parallel to the wire and is separated from it by the distance which is η=1.5\eta=1.5 times greater than the side of the frame. Find: (a) Ampere force acting on the frame; (b) the mechanical work to be performed in order to turn the frame through 180∘180^{\circ} about its axis, with the currents maintained constant

Reveal Answer
 3.255. (a) F=2μ0II0/π(4η2−1)=0.40μN; (b) A==(μ0aII0/π)ln⁡[(2η+1)/(2η−1)]=0.10μJ\begin{array}{l} \text { 3.255. (a) } F=2 \mu_{0} I I_{0} / \pi\left(4 \eta^{2}-1\right)=0.40 \mu N ; \quad \text { (b) } \quad A= \\ =\left(\mu_{0} a I I_{0} / \pi\right) \ln [(2 \eta+1) /(2 \eta-1)]=0.10 \mu \mathrm{J} \end{array}
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