A short magnet oscillates in an oscillation magnetometer with a time period of 0.10s where the earth's horizontal magnetic field is `24 muT`. A downward current of `18A` is established in a vertical wire placed 20cm east of the magnet. Find the new time period.

Options

A. `0.1 s`
B. `0.089 s`
C. `0.076 s`
D. `0.057 s`

Answer

C

Explanation

Initially `T=2pisqrt(I/(mB_(H)))`,
Finally `T^(')=2pisqrt(I/(m(B+B_(H))))`
Where `B=` Magnetic field due to down ward conductor
`=mu_(0)/(4pi).(2i)/a=18 muT`
`:. T^(')/T=sqrt(B_(H)/(B+B_(H)))impliesT^(')/0.1=24/(18+24)impliesT^(')=0.076 s`.