Question

A rectangular coil of wire of area $400c{m^2}$ contains 500 turns. It is placed in a magnetic field of induction $4 \times {10^{ - 3}}T$ and it makes an angle $60^\circ $ with the field. A current of 0.2A is passed through it. The torque on the coil is:
A) $8\sqrt 3 \times {10^{ - 3}}N - m$.
B) $8 \times {10^{ - 3}}N - m$.
C) $8\sqrt 3 \times {10^{ - 4}}N - m$.
D) $8 \times {10^{ - 4}}N - m$.

Answer 1

Hint:Magnetic field is defined as the influence of the magnetic field on the moving electric charges. The product of the force applied and the distance from the lever’s fulcrum is known as torque. The direction of the torque is decided by the direction of force and the direction vector.

Formula used:The formula of the torque due to magnetic field,
$ \Rightarrow T = \left( {NiAB} \right) \cdot \cos \theta $
Where torque is T the number of turns is N the current is I the area of cross section is A the magnetic field is B and the angle between the magnetic field and the coil is$\theta $.

Complete step by step solution:
It is given in the problem that a rectangular coil of wire of area $400c{m^2}$ contains 500 turns it is placed in a magnetic field of induction $4 \times {10^{ - 3}}T$ and it makes an angle $60^\circ $ with the field, a current of 0.2A is passed through the coil and we need to tell the torque produced on the coil.
Let us calculate the torque on the rectangular coil,
The formula of the torque due to magnetic field,
$ \Rightarrow T = \left( {NiAB} \right) \cdot \cos \theta $
Where torque is T the number of turns is N the current is I the area of cross section is A the magnetic field is B and the angle between the magnetic field and the coil is $\theta $.
The area of cross section is $400c{m^2}$ the number of turns are 500 the magnetic field of induction is $4 \times {10^{ - 3}}T$ the current through the coil is 0.2A and the angle between the coil and the magnetic field is equal to $60^\circ $_.
$ \Rightarrow T = \left( {NiAB} \right) \cdot \cos \theta $
$ \Rightarrow T = \left[ {\left( {500} \right) \times \left( {0 \cdot 2} \right) \times \left( {400 \times {{10}^{ - 4}}} \right) \times \left( {4 \times {{10}^{ - 3}}} \right)} \right] \cdot \cos 60^\circ $
$ \Rightarrow T = \left[ {\left( {500} \right) \times \left( {0 \cdot 2} \right) \times \left( {400 \times {{10}^{ - 4}}} \right) \times \left( {4 \times {{10}^{ - 3}}} \right)} \right] \cdot \left( {\dfrac{1}{2}} \right)$
$ \Rightarrow T = 8 \times {10^{ - 3}}N - m$.
The torque on the coil is equal to $T = 8 \times {10^{ - 3}}N - m$.

The correct answer is option B.

Note:The current in the coil creates a magnetic field around the coil. The coil has a magnetic field which opposes or attracts the magnetic field due to which there is generation of the torque. The torque due to the magnetic field depends on the angle between the coil and the magnetic field.