A thin uniform rod with negligible mass and l | vedclass

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Question

(A) The magnetic force on an element $dr$ at a distance $r$ from the hinge $P$ is $dF = i(dr)B$.The torque due to this magnetic force about the hinge

Vedclass · Accepted Answer

$\frac{5ilB}{8k}$

Vedclass · Answer

(A) The magnetic force on an element $dr$ at a distance $r$ from the hinge $P$ is $dF = i(dr)B$.The torque due to this magnetic force about the hinge $P$ is $d	au = (dF)r = (iB dr)r$.Integrating from $r = 0$ to $r = l$,the total torque is $	au = \int_{0}^{l} iBr dr = \frac{il^2B}{2}$.In the equilibrium position,the rod makes an angle of $53^{\circ}$ with the horizontal. The spring is attached to the top end of the rod. The extension in the spring is $x$. The spring force is $F_s = kx$.The torque due to the spring force about the hinge $P$ is $	au_s = (kx) 	imes (l \sin 53^{\circ})$.Equating the torques: $\frac{il^2B}{2} = (kx)(l \sin 53^{\circ})$.Since $\sin 53^{\circ} = \frac{4}{5}$,we have $\frac{il^2B}{2} = kxl(\frac{4}{5})$.Solving for $x$: $x = \frac{5ilB}{8k}$.

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