A magnet having time period 2 oscillations/s at a place having magnetic field intensity of `0.4 xx10^(-5)` T. At another place, it takes 2 s to complete one oscillation. The value of the earth's horizontal field at that place is ............`10^(-7)` T

Answer

2.5

Explanation

Time period,
`T=2pisqrt(l/(MB))`
So, `T prop 1/sqrtB`
So, `T_1/T_2=sqrt(B_2/B_1)`
Now `T_1 =1//2s=0.5s`
`T_2 = 2s`
so , `(0.5)/2=sqrt((B_2)/(4xx10^(-6)))`
`B_2 = (4xx10^(-6))/16`
`= 2.5 xx10^(-7)T`