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.

When the rod is in equilibrium and makes an angle of `53.0^@` with the floor, is the spring stretched or compressed?

Answer

A, C

Explanation

a. Yes, magnetic force for calculation of torque can beassumed at centre. Since variatonof torque about `P` from one end of the rod to the other end comes out to be linear.
`:. tau=(IlB)(l/2)=(Il^2B)/2`
`=((6.5)(0.2)^2(0.34))/2`
`=0.0442N-m`
b. Magnetic torque on rod will come out to be clockwise. Therfore, torque of spring force should be anticlockwise or spring should be stretched.
b. In equilibrium,
clockwise torque of magnetic force
`=`anticlockwise torque of spring force
`:. 0.0442=(kx)(lsin53^@)=(4.8)(x)(0.2)(4/5)`
`or x=0.057m`
`U=1/2kx^2=1/2xx(4.8)(0.057)^2`
`=7.8xx10^-3J`