Question

A circular loop of area $0.01{{m}^{2}}$ carrying a current of 10 A, is held perpendicular to a magnetic field
of intensity 0.1 T. The torque acting on the loop is
A. 0.001Nm
B. 0.8Nm
C. Zero
D. 0.01Nm

Answer 1

Hint:In this question all values are given to find torque acting on a current carrying loop placed in a magnetic field. We also know the direct equation to find torque. Here the angle between the magnetic field and area vector of the loop is to find out inorder to solve this problem. Remember that the area vector is always normal to the plane containing the loop.



Formula Used:
Torque acting on current carrying loop placed in a magnetic field, \[\tau =\vec{\mu }\times \vec{B}\]
Where µ is the magnetic moment and B is intensity of magnetic field
We have equation for magnetic moment as:
Magnetic moment, $\mu =IA$
Where I is the current flowing through the loop and A is the area of the loop.
Now we can rewrite the equation for torque acting on a loop as:
Torque, $\tau =IAB\sin \theta $
Where ϴ is the angle between magnetic field and area vector.



Complete answer:
We have a direct equation for torque acting on a loop.
Torque acting on a current carrying loop, $\tau =IAB\sin \theta $
From question, we have:
 Current carried by the loop, $I=10A$
Area of the current carrying loop, $A=0.01{{m}^{2}}$
Intensity of magnetic field, $B=0.1T$
And angle between magnetic field and area vector, $\theta =0{}^\circ $
Therefore, torque acting on the loop, $\tau =10\times 0.01\times 0.1\times \sin 0=0$

Thus, the correct option is C.



Note:Here the angle between magnetic field and area vector is zero since it is given in the question that the loop is held perpendicular to the magnetic field. That is, an area vector which is always perpendicular to the plane of the loop and magnetic field are now parallel to each other. Usually, the equation for torque contains a term N which is the number of turns. Here nothing is mentioned in the question, so take it as one.