The figure shows a conductor of weight `1.0 N&` length `L=0.5 m` placed on a rough inclined plane making an angle `30^(@)` with the horizontal so that conductor is perpendicular to a uniform horizontal magnetic field of induction `B=0.10 T`.The coefficient of static friction betwen the conductor inside the plane of is `0.1 A` A current of `l=10 A` flows through the conductor inside the plane of this paper as shown.What is the force needed to be applied parallel to the inclined plane to keep the conductor at rest?

Answer

`3/4(1-sqrt3/10)N "to"3/4(1+sqrt3/10)N`

Explanation

`N=(mg+i.lB)cos alpha`
`F_(max)=(mg+i.l.B)sin apha+muN`
`F_(max)=3/4(1+sqrt3/4)N`
`F_(min)=(mg+i.lB) sin alpha-muN`
`F_(min)3/4(1-sqrt3/4)N`