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=4.80Nm^-1` connects the other end of the rod to a vertical wall. 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=6.50 A in the rod, in the direction shown.

Calculate the torque due to the magnetic force on the rod, for an axis at P.
A. `0.0442Nm^-1`, clockwise
B. `0.0442Nm^-1`, anticlockwise
C. `0.022Nm^-1`, clockwise
D. `0.022Nm^-1`, anticlockwise

Answer 1

Correct Answer - a
By exmining a small piece of the wire Fig. we find:

Dividing the rod into infinitiesimal sections of length dr.
The magnetic force on this section is `dF_1=IB dr` and is
perpendicular to the rod. The torque dt due to forceon this section
is `dtau=rdF_1=IBrdr`
The total torque is
`intdr=IBint_0^1r dr=1/2Il^2B=0.0442Nm^-1`,
clockwise. This is the same torque calculated from a force diagram
in which the total magnetic force `F_1 =ILB` acts at the centre of
the rod.
`F_1` produces a clockwise torque, so the spring force must produce
a couterclockwise torque. The spring force must be to left,
the spring is stretched:
axis at hinge, counterclockwise torque positive
`(kx)lsin 53^@-1/2Il^2B=0`
`x=(IlB)/(2ksin 53.0^@)=((6.50A)(0.200m)(0.340T))/(2(4.80Nm^-1) sin 53.0^@)`
`=0.05765m`