A horizontal wire of length 0.5m carries...

A horizontal wire of length `0.5m` carries a current of `5A`. If the mass of the wire is `10mg`, the minimum magnetic field required to support the weight of the wire is `(g=10m//g^(2))`

`4 xx 10^(-4) T`

`25 xx 10^(-4) T`

`4 xx 10^(-1) T`

`25 xx 10^(-1) T`

Verified by Experts

The correct Answer is:

`B = (mg)/(il)`

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A horizontal conducting wire of length 0.1 m carries a current of 4A. If mass of the wire is 7.5 xx 10^(-5) kg and g=10 m//s^(2) , the uniform magnetic induction required to support the weight of the wire is

`1.675 xx 10^(-3)` T

`1.775 xx 10^(-3)` T

`1.875 xx 10^(-3)` T

`1.975 xx 10^(-3)` T

`6xx10^-3T`, acting vertically upwards

`6xx10^-3T`, acting horizontally perpendicular to wire

`6xx10^-2T`, acting vertically downwards

`6xx10^-2T`, acting horizontally perpendicular to wire

2.4 N

1.2 N

1.6 N

3.2 N