A 100 turn rectangular coil ABCD (in XY plane) is hung from one arm of a balance (figure). A mass `500g` is added to the other arm to balance the weight of the coil. A current `4*9A` passes through the coil and a constant magnetic field of `0*2T` acting inward (in xz plane) is switched on such that only arm CD of length `1cm` lies in the field. How much additional mass 'm' must be added to regain the balance?

Options

A. `1kg`
B. `1gm`
C. `2kg`
D. `2gm`

Answer

2

Explanation

Here, `M=500g=0.5kg,B=0.2T`
When the magnetic field is absent
`sumtau=0`
`Mgl=W_(coil)limpliesW_(coil)=Mg`
`:. W_(coil)=0.5xx9.8=4.9N`
When current `I` is passed throught the coil and the magnetic field is switched on, let `m` be the mass added in a pan to balance the beam.
Then
`Sigmatau'=0`
`Mgl+mgl=W_(coil)l+(IBLsin90^(@))xxl`
`:. m=(0.2xx4.9xx1xx10^(-2))/(9.8)=10^(-3)kg=1g`