Shown a short magnet executing small oscillations in vibration magnetometers in earth's magnetic field having horizontal component `24 muT`. The period of oscillation is `0.1 s`. When the key `K` is closed, an upward current of `18 A` is established as shown. The new time period is

Options

A. `0.1 s`
B. `0.2 s`
C. `0.3 s`
D. `0.4 s`

Answer

B

Explanation

`B=(mu_(0)I)/(2pir)=(4pixx10^(-7)xx18)/(2pixx0.2)T=18 muT`
Now, `T=2pisqrt(I/(MB_(H)))` and `T^(')=2pisqrt(I/(M(B_(H)-B)))`
Dividing
`T^(')/Tsqrt(B_(H)/(B_(H)-B)) or T^(')/T=sqrt(24/(24-18))=2`
`T^(')=2xx0.1 s=0.2 s.`