A straight wore of mass 2 gram and length 50 cm is kept horizontal in a uniform magnetic field on induction `2xx 10^(-2)Wb//m^(2).` The field is horizontal and is perpendicular to the length of the wire. How much current should be passed through wire, so to balance its weight ?
`(g=10m//s^(2))`

Options

A. `1A`
B. `1.5`A
C. `2A`
D. `2.5` A

Answer

C

Explanation

Given: `m=2g =2xx10^(-3) kg, L =0.5m,`
`B = 2xx10 ^(-2) Wb//m^(2) and g=10m//s^(2)`
The force F acting on the wire fue to magnetic induction is `F =BIL.`
For balancing the weight, the force be equal to the weigth (mg) of the wire.
`therefore F =BIL=mg`
`therefore I =(mg)/(BL) =(2xx10^(-3) xx10)/(2xx10^(-2) xx0.5) =2A`