<p>A cube, 40.0 cm on a side, is placed in a uniform magnetic field of magnitude

Question

<p>A cube, 40.0 cm on a side, is placed in a uniform magnetic field of magnitude B = 20.0 mT as shown. Four straight segments of wire, ab, bc, cd, and da, form a closed conductive path that carries a current I = 5.00 A. Explain your reasoning for the key steps when you: <br />(a) determine the magnitude and direction of the force acting on each of the four current segments; and <br />(b) show the total force acting on the completed circuit path sums to zero.</p>
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Explanation / Answer

Given Side of the cube , l = 0.4 m Magnitude of magnetic field , B = 0.02 T Magnitude of current is , I = 5.0 A ab = bc = 0.4 m cd = da = l2 = 0.4m 2 = 0.56 m a) For ab segment:     Magnetic force , F1 = 0 ,as B and I are in opposite directions    For bc segment:     F2 = IBl sin90 = 5.0A* 0.02T*(0.4 m) = 0.04N     I is along +Z axis , B is along +Y axis .Direction of F is along     - X axis    For cd segment :     F3 = IBl sin45 = 5.0A* 0.02T*(0.56 m)(0.707) = 0.04N    Direction of F is along - Z axis    For da segment :     F4 = IBl sin90 = 5.0A* 0.02T*(0.56 m)(1) = 0.056N    Direction of F is parallel to X -Z plane, at 45^0 with +X and +Z direction b) Net force acting is      F = F1 + F2 +F3 +F4         = 0 + 0.04 N (-i) + (0.04N) (-k) +0.056 cos45 (+i)+ 0.056 sin45(+k)    F = 0 Hence, proved     F4 = IBl sin90 = 5.0A* 0.02T*(0.56 m)(1) = 0.056N    Direction of F is parallel to X -Z plane, at 45^0 with +X and +Z direction b) Net force acting is      F = F1 + F2 +F3 +F4         = 0 + 0.04 N (-i) + (0.04N) (-k) +0.056 cos45 (+i)+ 0.056 sin45(+k)    F = 0 Hence, proved

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