The earth's magnetic field at the geometric poles is `sqrt(10)xx10^(-5)` T. The magnitude of the field at a point on the earth's surface where the radius makes an angle `theta` with the axis of earth's assumed magnetic dipole is `5xx10^(-5)` T. The magnitude of `theta` in degree is :

Options

A. `30^(@)`
B. `60^(@)`
C. `45^(@)`
D. `75^(@)`

Answer

C

Explanation

According to the question,
`B_(v)=sqrt(10)xx10^(-5)T`

In `DeltaOBC`,
`:.cos theta=(B_(v))/(B_(R))=(sqrt(10)xx10^(-5))/(5xx10^(-5))=(sqrt(2))/(sqrt(5))`
or `theta=50.76^(@)~~45^(@)`
Hence, the magnitude of `theta` is `45^(@)`.