Question

A short oscillates in an oscillation magnetometer with a time period of $0.10s$ where the earth's horizontal magnetic field $24\mu T$.$A$ downward current of $18A$ is established in a vertical wire placed $20cm$ east of the magnet. Find the new time period

Answer 1

A short oscillates in an oscillation magnetometer with a time period of$=0.10s$

where the earth's horizontal magnetic field $=24\mu T$

A downward current $=18A$

Vertical placed$=20cm$east of the magnet

Find the new time period

Solution

Here,

Horizontal component of earth's magnetic field

$B_H=24\times {10}^{-6}T$

Time period of oscillation

$T_1=0.1s$

Downward current in the vertical wire$I=18A$

So, $T=2\pi \sqrt{\frac{I}{MB}}$ base H here

$g^{\prime }ITbase1=\frac{1}{40}min{T}_{base2}=?\frac{T_{base1}}{T_{base2}}=\sqrt{\frac{i}{T}}=\frac{1}{140}Tbase2=\sqrt{1/2}=1/600Tbase{2}^2=1/2=T_{base2}^2=1/800T_{base2}=0.03536min$

For 1 oscillation time taken $=0.3536min$

For $40$ oscillation time$=4\times 0.03536=1.414=\sqrt{2}min$