A straight rod of mass m and length L is susp | vedclass

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(C) $1$. Initially,the rod is in equilibrium under the force of gravity and the spring force. Let $k$ be the spring constant of each spring. The total

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$\frac{mgR}{\varepsilon L}$; directed into the plane of the paper

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(C) $1$. Initially,the rod is in equilibrium under the force of gravity and the spring force. Let $k$ be the spring constant of each spring. The total upward force from two springs is $2kx_0 = mg$,so $k = \frac{mg}{2x_0}$.$2$. When the magnetic field $B$ is switched on,a magnetic force $F_m = ILB$ acts on the rod. The current in the rod is $I = \frac{\varepsilon}{R}$.$3$. The springs extend by an additional distance $x_0$,meaning the total extension is $2x_0$. The new equilibrium condition is $2k(2x_0) = mg + F_m$.$4$. Substituting $2kx_0 = mg$ into the equation: $2(2kx_0) = 2mg = mg + F_m$,which implies $F_m = mg$.$5$. Since $F_m = ILB = \frac{\varepsilon}{R}LB = mg$,we solve for $B$: $B = \frac{mgR}{\varepsilon L}$.$6$. To increase the extension,the magnetic force must act downwards. By Fleming's Left-Hand Rule,with current flowing in a specific direction,the magnetic field must be directed into the plane of the paper to produce a downward force.

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