Question

A thin, uniform rod with negligible mass and length $0.200m$ is attached to the floor by a frictionless hinge at point $P($as shown in fig$)$. A horizontal spring with force constant $k=4.80N{m}^{-1}$ connects the other end of the rod to a vertical wall. The rod is in a uniform magnetic field $B=0.340T$ directed into the plane of the figure. There is current $I=6.50A$ in the rod, in the direction shown. Calculate the torque due to the magnetic force on the rod, for an axis at P:

• A. $0.0442N{m}^{-1}$, clockwise
• B. $0.0442N{m}^{-1}$, anitclockwise
• C. $0.022N{m}^{-1}$, clockwise
• D. $0.022N{m}^{-1}$, anticlockwise

Answer 1

The correct option is A $0.0442N{m}^{-1}$, clockwise

Divide the rod into infinitesimal segments of length dr.
Consider the segment of length dr at a distance r from point P( hinge).
Force on dr $dF=iBdr$
Torque on dr$d\tau =rdF=iBrdr$

Total torque on the ord of length $l\tau ={\int }_0^{l}iBrdr=\frac{iB{l}^2}{2}$

Given, $i=6.5A$, $B=0.34T$;$l=0.2m$

$\therefore \tau =\frac{6.5\times 0.34\times (0.2{)}^2}{2}=0.0442Nm$