A straight conductor of length `60cm` and mass `10gm` is suspended horizontally by a pair of flexible leads in a magnetic field of `0.4T` as shown in figure. A current at `0.41A` should be passed through the conductor to just remove the tension in the supporting cable. The total tension in the leads if the direction of current is reversed is found `1.96xx10^(-y)N`. Find `y` ?

Answer

1

Explanation

In order to just remove the tension in the leads, the weight of the conductor must be balanced by or Bil = mg
`i=(md)/(Bl)=(10xx10^(-3)xx9.8)/(0.4xx0.6)" "i=0.41A`
In second case mg and iBl both will act down ward.
Therefore tension in the cable is `2mg=1.96xx10^(-1)N`