Question

A straight conductor of weight $1N$ and length $0.5m$, is located in a place making an angle ${30}^o$ with the horizontal so that it is perpendicular to a uniform magnetic field of induction $B=0.109T$.

Given that the conductor carries a current of $10A$ away from the reader, and coefficient of static friction is $0.1$, find the force needed to be applied parallel to the plane to sustain the conductor at rest.

• A. 0.63
• B. 0.9
• C. between 0.63 to 0.90
• D. None of the above

Answer 1

The correct option is C between 0.63 to 0.90

For conductor, just to prevent from moving downward:
$(mg+Bil)sin\theta =P_1+\mu (mg+Bil)cos\theta$

$P_1=(mg+Bil)(sin\theta -\mu cos\theta )$
$=(1+0.109\times 10\times 0.5)\times (sin30-0.1\times cos30)$
$=0.77-0.13=0.63N$
For the conductor to just move upward, here friction will act in downward direction, then net force required to move the conductor up is:

$P_2=(mg+Bil)(sin\theta +\mu cos\theta )$
$=0.77+0.13=0.90N$
Hence, the force should be between 0.63 N and 0.90 N.