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(A) In a deflection magnetometer,the magnetic field due to the magnet $F$ and the horizontal component of the earth's magnetic field $B_H$ are perpend

Vedclass · Accepted Answer

$T^2=T_0^2 \cos 	heta$

Vedclass · Answer

(A) In a deflection magnetometer,the magnetic field due to the magnet $F$ and the horizontal component of the earth's magnetic field $B_H$ are perpendicular to each other.Therefore,the net magnetic field acting on the needle is $B_{net} = \sqrt{F^2 + B_H^2}$.The time period of oscillation is given by $T = 2\pi \sqrt{\frac{I}{m B_{net}}}$,where $I$ is the moment of inertia and $m$ is the magnetic moment of the needle.Thus,$T = 2\pi \sqrt{\frac{I}{m \sqrt{F^2 + B_H^2}}}$.When the magnet is removed,the field acting on the needle is only $B_H$,so the time period is $T_0 = 2\pi \sqrt{\frac{I}{m B_H}}$.From the principle of the deflection magnetometer,we have $\frac{F}{B_H} = 	an 	heta$,which implies $F = B_H 	an 	heta$.Substituting $F$ into the expression for $T$:$T = 2\pi \sqrt{\frac{I}{m \sqrt{(B_H 	an 	heta)^2 + B_H^2}}} = 2\pi \sqrt{\frac{I}{m B_H \sqrt{	an^2 	heta + 1}}} = 2\pi \sqrt{\frac{I}{m B_H \sec 	heta}}$.Comparing $T$ and $T_0$:$\frac{T}{T_0} = \sqrt{\frac{1}{\sec 	heta}} = \sqrt{\cos 	heta}$.Squaring both sides,we get $T^2 = T_0^2 \cos 	heta$.

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