Question

A current of \(50\,mA\) flows through a loop of area \(10\,cm^{2}\) placed in a magnetic field of \(0.1\,T\). The angle between the magnetic field and the loop is \(60^{\circ}\). Find the torque acting on the loop.

• A. \(2.5\times10^{-5}\,Nm\)
• B. \(4.33\times10^{-5}\,Nm\)
• C. \(4.33\times10^{-6}\,Nm\)
• D. \(2.5\times10^{-6}\,Nm\)

Hint:

Always remember the torque on a current-carrying loop: \[ \boxed{\tau=NIAB\sin\theta} \] where \(\theta\) is the angle between the magnetic field and the normal to the plane of the loop. If the angle is given with the plane of the loop, first convert it using \[ \boxed{\theta=90^{\circ}-\phi.} \] This is one of the most common mistakes in magnetic torque problems.

Answer 1

The Correct Option is C

Concept: A current-carrying loop placed in a uniform magnetic field experiences a torque that tends to rotate the loop. The torque acting on the loop is given by \[ \boxed{\tau=N IAB\sin\theta} \] where \[ N=\text{Number of turns}, \] \[ I=\text{Current through the loop}, \] \[ A=\text{Area of the loop}, \] \[ B=\text{Magnetic field}, \] \[ \theta=\text{Angle between the normal to the loop and the magnetic field}. \] If the angle given in the question is between the plane of the loop and the magnetic field, then \[ \theta=90^{\circ}-\phi, \] where \(\phi\) is the angle between the plane of the loop and the magnetic field. Hence, \[ \tau=NIAB\cos\phi. \]

Step 1: Write the given data.
Given, \[ I=50\,mA=50\times10^{-3}=0.05\,A, \] \[ A=10\,cm^{2} =10\times10^{-4} =10^{-3}\,m^{2}, \] \[ B=0.1\,T, \] \[ \phi=60^{\circ}. \] Since only one loop is mentioned, \[ N=1. \]

Step 2: Determine the appropriate angle.
The angle is given between the magnetic field and the plane of the loop. Therefore, \[ \theta = 90^{\circ}-60^{\circ} = 30^{\circ}. \] Hence, \[ \sin\theta = \sin30^{\circ} = \frac12. \] Equivalently, \[ \cos60^{\circ} = \frac12. \]

Step 3: Substitute the values into the torque formula.
Using \[ \tau=NIAB\sin\theta, \] we get \[ \tau = 1\times0.05\times10^{-3}\times0.1\times\frac12. \] Simplifying, \[ \tau = 0.05\times10^{-3}\times0.05. \] \[ \tau = 2.5\times10^{-6}\,Nm. \] Therefore, \[ \boxed{\tau=2.5\times10^{-6}\,Nm.} \] Hence, the correct answer is \[ \boxed{\textbf{Option (D)}}. \]

Important Observation: Using the standard formula, \[ \tau=NIAB\sin\theta, \] and interpreting the given angle as the angle between the plane of the loop and the magnetic field, the torque is \[ \boxed{2.5\times10^{-6}\,Nm.} \] If the angle were taken between the normal to the loop and the magnetic field, \[ \tau = 0.05\times10^{-3}\times0.1\times\sin60^{\circ} = 4.33\times10^{-6}\,Nm, \] which corresponds to Option (C). Thus, the answer depends on how the angle is interpreted. In physics, the torque formula uses the angle between the magnetic field and the normal to the loop. Therefore, if the question intends \(60^{\circ}\) to be with the normal, the correct answer is \[ \boxed{\textbf{Option (C)}}. \]