Question

A magnetic needle suspended parallel to a magnetic field requires \[\sqrt 3 \,{\text{J}}\] of work to turn it through \[{60^ \circ }\]. The torque needed to maintain the needle in this position will be:-
A. \[3\,{\text{J}}\]
B. \[\sqrt 3 \,{\text{J}}\]
C. \[\dfrac{3}{2}\,{\text{J}}\]
D. \[2\sqrt 3 \,{\text{J}}\]

Answer 1

Hint:Use the formula for work done in changing the orientation of a magnetic substance in the presence of a magnetic field. Calculate the value and equate this value with the given value of work done to find the magnetic field and magnetic moment of the needle. Then use these values to find the required torque.

Complete step by step answer:
Given, the magnetic needle is turned through \[{60^ \circ }\]which means that initial angle \[{\theta _1} = {0^ \circ }\] and the final angle \[{\theta _2} = {60^ \circ }\]. Work required to turn the magnetic needle, \[W = \sqrt 3 \,{\text{J}}\].

We have the formula for work done in changing the orientation of a magnetic substance from \[{\theta _1}\] to \[{\theta _2}\] in a magnetic field as,
\[W = MB(\cos {\theta _1} - \cos {\theta _2})\], where \[M\] is the magnetic moment of substance, \[B\] is the magnetic field.

Now, putting the values of \[{\theta _1}\] and \[{\theta _2}\] in the above expression we get
\[W = MB(\cos {0^ \circ } - \cos {60^ \circ }) \\
\Rightarrow W = MB\left( {1 - \dfrac{1}{2}} \right) \]
\[ \Rightarrow W = \dfrac{{MB}}{2}\] (i)
Formula for torque is,
\[\overrightarrow \tau = \overrightarrow M \times \overrightarrow B \\
\Rightarrow \tau = MB\sin \theta \]
The needle should be maintain at \[{60^ \circ }\], so the required torque will be
\[\tau = MB\sin {60^ \circ }\]
\[ \Rightarrow \tau = MB\dfrac{{\sqrt 3 }}{2}\] (ii)
Diving equation (ii) by (i), we get
\[\dfrac{\tau }{W} = MB\dfrac{{\sqrt 3 }}{2} \times \dfrac{2}{{MB}} \\
\Rightarrow \dfrac{\tau }{W} = \sqrt 3 \\
\Rightarrow \tau = \sqrt 3 W \]
The work done is given as\[\sqrt 3 \,{\text{J}}\]. Putting substituting this value of \[W\] in the above equation, we get
\[\therefore \tau = \sqrt 3 \sqrt 3 = 3\,{\text{J}}\]
Therefore, the torque required to maintain the needle at \[{60^ \circ }\] is \[3\,{\text{J}}\]

Hence, the correct answer is option A.

Note:The magnetic potential energy of the needle is given by \[U = \overrightarrow M \cdot \overrightarrow B = - MB\cos \theta \]. At \[\theta = {0^ \circ }\], the potential energy is \[U = - MB\] and at \[\theta = 18{0^ \circ }\] the potential energy is \[U = MB\]. This implies that when the magnetic needle is parallel to the direction of magnetic field, the potential energy is minimum and it is the most stable position of the needle and when the needle is anti-parallel to the direction of magnetic field, the potential energy is maximum and it is the most unstable position of the needle.