Question

T0 is time period of oscillation of the needle in uniform horizontal magnetic field (B h) of earth. If another magnetic field is applied perpendicular to B, such that the needle deflects by theta for its equilibrium. Now period of oscillation of needle is T. The relation between Tand T0 id​

Answer 1

Answer:

Hind and explations are given below

Explanation:

Answer 2

Given:

In the usual setting of deflection magnetometer, filled due to magnet(F)and horizontal component(H)of earth field are perpendicular to each other.

To find:

Find the Relation between $T$ and $T^{0}$ .

Explanation:

Net field on the magnetic needle is,

$\sqrt{F^{2}+H^{2}}$

When the magnet  is removed,

$T_{0}=2\pi \sqrt{\frac{I}{MH} }$

$T=2\pi \sqrt\frac{I}{M\sqrt{F^{2}+H^{2}} }$

$\frac{T}{T_{0}}=\frac{2\pi \sqrt\frac{I}{M\sqrt{F^{2}+H^{2}} }}{2\pi \sqrt{\frac{I}{MH} }}$

$=\sqrt{\frac{H}{H\sqrt{secc^{2}0} } }$

$\sqrt{cos0}$

$T^{2}=T_{0}^{2}cos$∅

Answer:

Therefore, the relation between $T$ and $T_{0}$ the $T^{2}=T_{0}^{2}cos$∅.