A thin uniform rod with negligible mass and l | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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 $x$: $x = \frac{5ilB}{8k}$.

Similar Questions

Currents of $10\, A$ and $2\, A$ are passed through two parallel thin wires $A$ and $B$ respectively in opposite directions. Wire $A$ is infinitely long and the length of wire $B$ is $2\, m$. The force acting on the conductor $B$,which is situated at a distance of $10\, cm$ from $A$,will be:

$A$ uniform conducting wire $ABC$ has a mass of $10 \, g$. $A$ current of $2 \, A$ flows through it. The wire is kept in a uniform magnetic field $B = 2 \, T$. The acceleration of the wire will be ............. $m \, s^{-2}$.

If two streams of protons move parallel to each other in the same direction,then they

$A$ current-carrying rectangular loop is placed near a straight infinitely long current-carrying wire as shown in the figure. The torque acting on the loop is

Vedclass Test Series

Exam Paper Generator

Online Exam Module