A horizontal metallic rod of mass m and lengt | vedclass

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Question

(B) The forces acting on the rod are the gravitational force $mg$ (downwards) and the magnetic force $F_m = ilB$ (upwards,by Fleming's Left-Hand Rule)

Vedclass · Accepted Answer

$l_0 + \frac{ilB - mg}{2K}$

Vedclass · Answer

(B) The forces acting on the rod are the gravitational force $mg$ (downwards) and the magnetic force $F_m = ilB$ (upwards,by Fleming's Left-Hand Rule).The net force on the rod is $F_{net} = ilB - mg$ (upwards).Since the rod is in equilibrium,the two springs must provide a restoring force equal to the net force. Let $x$ be the extension in each spring. The total restoring force provided by the two springs is $2Kx$.Equating the forces: $2Kx = ilB - mg$.Solving for the extension $x$: $x = \frac{ilB - mg}{2K}$.The length of each spring in the equilibrium state is $l = l_0 + x$.Substituting the value of $x$: $l = l_0 + \frac{ilB - mg}{2K}$.

$A$ horizontal metallic rod of mass $m$ and length $l$ is supported by two vertical identical springs of spring constant $K$ each and natural length $l_0$. $A$ current $i$ is flowing in the rod in the direction shown. If the rod is in equilibrium,then the length of each spring in this state is:

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