Question

A coil in the shape of an equilateral triangle of side 10 cm lies in a vertical plane between the pole pieces of permanent magnet producing a horizontal magnetic field 20 mT. The torque acting on the coil when a current of 0.2 A is passed through it and its plane becomes parallel to the magnetic field will be \( \sqrt{x} \times 10^{-5} \) Nm. The value of x is __________.

Hint:

Remember that the angle \(\theta\) in the torque formula \( \tau = MB \sin\theta \) is between the magnetic moment vector (normal to the coil's area) and the magnetic field vector. When the plane of the coil is parallel to the field, \(\theta = 90^\circ\) and the torque is maximum. When the plane is perpendicular to the field, \(\theta = 0^\circ\) and the torque is zero.

Answer 1

Correct Answer: 3

Step 1: Understanding the Question:
We need to find the torque on a triangular current-carrying coil placed in a magnetic field. The orientation is specified, which allows us to find the maximum torque.
Step 2: Key Formula or Approach:
1. The torque (\(\tau\)) on a current loop in a uniform magnetic field (B) is given by \( \vec{\tau} = \vec{M} \times \vec{B} \), where \(\vec{M}\) is the magnetic dipole moment of the loop.
2. The magnitude of the torque is \( \tau = MB \sin\theta \), where \(\theta\) is the angle between \(\vec{M}\) and \(\vec{B}\).
3. The magnetic moment is given by \( M = NIA \), where N is the number of turns (here N=1), I is the current, and A is the area of the loop.
4. Torque is maximum when \( \sin\theta = 1 \), which occurs when \(\vec{M}\) is perpendicular to \(\vec{B}\). The direction of \(\vec{M}\) is normal to the plane of the coil. If the plane of the coil is parallel to \(\vec{B}\), then its normal is perpendicular to \(\vec{B}\), so \(\theta=90^\circ\).
Step 3: Detailed Explanation:
Given values:
Side of equilateral triangle, a = 10 cm = 0.1 m
Magnetic field, B = 20 mT = 20 \( \times \) 10\(^{-3}\) T
Current, I = 0.2 A
Number of turns, N = 1
First, calculate the area (A) of the equilateral triangle coil:
\[ A = \frac{\sqrt{3}}{4} a^2 = \frac{\sqrt{3}}{4} (0.1 \text{ m})^2 = \frac{\sqrt{3}}{4} (0.01) \text{ m}^2 \] The plane of the coil is parallel to the magnetic field, so the angle \(\theta\) between the magnetic moment (normal to the plane) and the magnetic field is 90\(^\circ\). Thus, the torque is maximum.
\[ \tau = \tau_{max} = NIAB \] Substitute the values:
\[ \tau = (1) \times (0.2 \text{ A}) \times \left( \frac{\sqrt{3}}{4} \times 0.01 \text{ m}^2 \right) \times (20 \times 10^{-3} \text{ T}) \] \[ \tau = 0.2 \times \frac{\sqrt{3}}{4} \times 0.01 \times 20 \times 10^{-3} \] \[ \tau = (0.2 \times 20) \times \frac{\sqrt{3}}{4} \times 0.01 \times 10^{-3} \] \[ \tau = 4 \times \frac{\sqrt{3}}{4} \times 10^{-2} \times 10^{-3} \] \[ \tau = \sqrt{3} \times 10^{-5} \text{ Nm} \] The problem states that the torque is \( \sqrt{x} \times 10^{-5} \) Nm.
Comparing our result with the given expression:
\[ \sqrt{3} \times 10^{-5} = \sqrt{x} \times 10^{-5} \] \[ \sqrt{3} = \sqrt{x} \implies x = 3 \] Step 4: Final Answer:
The value of x is 3.