Question

A magnet oscillating in a horizontal plane has a time period of 2 second at a place where angle of dip is $30{} ^\circ $ and 3 seconds at another place where the angle of dip is $60{} ^\circ $. The ratio of resultant magnetic fields at the two places is
(A) $\dfrac {4\sqrt {3}} {7} $
(B) $\dfrac {4} {9\sqrt {3}} $
(C) $\dfrac {9} {4\sqrt {3}} $
(D) $\dfrac {9} {\sqrt{3}}$

Answer 1

Hint: In this given question, we will use the formula of time period of oscillating magnet and observe how T is dependent on horizontal magnet field. And then, we will write two expressions of time period for given data and analyze them which as a result will give us a ratio resultant of the magnetic field.

Complete step-by-step answer:
We have been given that –
 T1= 2 second
T2 = 3 second
${{\delta} _ {1}} =30{} ^\circ $
${{\delta} _ {2}} =60{} ^\circ $
We are going to use the formula of time period of oscillation of magnet,
$T=2\Pi \sqrt{\dfrac{I}{M\times {{B} _ {H}}}} $
  Where, T = time period,
 I = moment of inertia
M = magnetic moment,
BH = horizontal magnetic field.
Now, we know for the expression of time period of oscillating magnet that -
$T\propto \sqrt{\dfrac{1}{{{B}_{H}}}}$
And, the horizontal component of magnetic field BH is given by B$\cos \delta $.
So, the two different expression of time period of oscillating magnet can be written as -
${{T} _ {1}}\propto \dfrac{1}{\sqrt{{{B}_{1}}\cos {{\delta} _ {1}}}} $ and,
${{T} _ {2}}\propto \dfrac{1}{\sqrt{{{B}_{2}}\cos {{\delta} _ {2}}}} $
Dividing T1 $\Rightarrow $ T2, we get:
$\Rightarrow \dfrac{{{T}_{1}}}{{{T}_{2}}}=\sqrt{\dfrac{{{B}_{2}}\cos {{\delta} _ {2}}} {{{B} _ {1}}\cos {{\delta} _ {1}}}} $
$\Rightarrow \dfrac{{{T}_{1}}^{2}}{{{T}_{2}}^{2}}=\dfrac{{{B}_{2}}\cos {{\delta} _ {2}}} {{{B} _ {1}}\cos {{\delta} _ {1}}} $
Or, $\Rightarrow \dfrac{{{B}_{1}}}{{{B}_{2}}}=\dfrac{{{T}_{2}}^{2}}{{{T}_{1}}^{2}}\times \dfrac{\cos {{\delta} _ {2}}} {\cos {{\delta} _ {1}}} $
$\Rightarrow \dfrac{{{B}_{1}}}{{{B}_{2}}}=\dfrac{{{3}^{2}}}{{{2}^{2}}}\times \dfrac{\cos 60{} ^\circ} {\cos 30{} ^\circ} $
$\Rightarrow \dfrac{{{B}_{1}}}{{{B}_{2}}}=\dfrac{9}{4}\times \dfrac{\dfrac{1}{2}}{\dfrac{\sqrt{3}}{2}}$
   $\therefore \dfrac{{{B}_{1}}}{{{B}_{2}}}=\dfrac{9}{4\sqrt{3}}$

So, the correct answer is “Option C”.

Note: If angle of dip is given in the question then relate it straight from the horizontal component of the magnetic field. And always keep in mind the expression of the time period of bar magnet oscillation then only you can relate the data given to you for finding the values of the magnetic field. Magnetic field has two components i.e. vertical components as well as horizontal components.