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$$ SERPSET$929.$P$\mathrm{.}044.$

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In the figure below, the cube is $39\mathrm{.}2$ cm on each edge. Four straight segments of wire$-$ab$,$ bc$,cd,$ and $da-$ form a closed loop that carries a current $I=4\mathrm{.}70A$ in the direction shown. A uniform magnetic field of magnitude $B=0\mathrm{.}019T$ is in the positive $y$ direction. Determine the magnetic force on each segment.

$($a$)ab$

$F=$ mN

$($b$)bc$

$F=$ mN

$($c$)cd$

$F=$ mN

$($d$)$ da

$F=$ mN

$($e$)$ Explain how you could find the force exerted on the fourth of these segments from the forces on the other three, without further calculation involving the magnetic field.

Since the magnetic forces on segments $ab,bc,$ and $cd$ are known, the force on da can be determined by the requirement that the net force on the entire closed loop is zero. This follows from Newton?s third law and the symmetry of the setup, as the loop is in equilibrium.

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