Passage
A thin uniform rod of negligible mass and length 0.200 m is attached to the floor by a frictionless hinge at point Pas shown in the figure. A horizontal spring with force contant `k=4.80Nm^(-1)` connects to the other end of the rod to a vertical wall. The rod is in a uniform magnetic field B=0.340 T 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.

Options

A. `0.0442 Nm^(-1)`, Clockwise
B. `0.0442Nm^(-1)`, Anticlockwise
C. `0.00442Nm^(-1)`, Clockwise
D. `0.00442Nm^(-1)`, Anticlockwise

Answer

A

Explanation

By examining a small piece of the wire we find
Divide the rod into infinitesimal sections oflength dr
The magnetic force on this section is `dF_(1) =IB dr` and is perpendicular to the rod. The torque due to force on this section is `dz=rdF_(1)=IBrdr`
The total torque is `int dz =I_(B) int_(0)^(l)rdr=(1)/(2)Il^(2)B=0.0442Nm^(-1)`