A wire of `62.0 cm` length and `13.0 g` mass is suspended by a pair of flexible leads in a uniform magnetic field of magnitude `0.440 T` in figure. What are the magnitude and direction of the current required to remove the tension in the supporting leads? Take `g = 10 m/s^2`

Answer

A, D

Explanation

`W=F_m` or `mg =ilB sin 90^@`
`:. i=i(mg)/(Bl)=((13x10^-3)(10))/(0.44xxx0.62)`
`=0.47A`
Magnetic force should be upwards to balance the weight. Hence from Fleming's left hnd rule we can se that direction of current should be from left to right.