Question

A thin uniform rod with negligible mass and length $l$ is attached to the floor by a frictionless hinge at point $P$. A horizontal spring with force constant $k$ connects the other end to wall. The rod is in a uniform magnetic field $B$ directed into the plane of paper. The extension in spring in equilibrium when a current is passed through the rod in direction shown is $\frac{ilB}{Nk}$. Find the value of $10N$. (Assuming spring to be in natural length initially.)

Answer 1

8

Answer 2

1. Magnetic Force: The magnetic force on the current-carrying rod is $\vec{F}_B = i(\vec{l} \times \vec{B})$. Given current $i$ is downwards along the rod and magnetic field $\vec{B}$ is into the plane, the force $\vec{F}_B$ is perpendicular to the rod, directed to the right (refer to the figure). The magnitude is $F_B = ilB \sin(90^\circ) = ilB$.
2. Spring Force: The spring is extended by $x$, so the spring force is $F_S = kx$. This force acts horizontally to the left.
3. Torque Balance: The rod is in equilibrium, so the net torque about the hinge P is zero.
   • Torque due to magnetic force $\tau_B$: The force $F_B$ acts perpendicular to the rod at its end (or effectively, the torque is $F_B \times l$). So, $\tau_B = F_B \times l = (ilB)l = il^2B$. This torque tends to rotate the rod clockwise.
   • Torque due to spring force $\tau_S$: The spring force $F_S$ acts horizontally. The perpendicular distance from the hinge P to the line of action of $F_S$ is $l \sin(53^\circ)$. So, $\tau_S = F_S \times l \sin(53^\circ) = kx l \sin(53^\circ)$. This torque tends to rotate the rod counter-clockwise.
4. Equilibrium Equation: $\tau_B = \tau_S$ $il^2B = kx l \sin(53^\circ)$ $ilB = kx \sin(53^\circ)$ Substitute $\sin(53^\circ) = 4/5$: $ilB = kx (4/5)$ $x = \frac{5ilB}{4k}$
5. Find N: The given extension is $x = \frac{ilB}{Nk}$. Comparing the two expressions for $x$: $\frac{5ilB}{4k} = \frac{ilB}{Nk}$ $\frac{5}{4} = \frac{1}{N} \implies N = \frac{4}{5} = 0.8$
6. Calculate 10N: $10N = 10 \times 0.8 = 8$.