Question

A circular coil of radius 2 cm and 125 turns carries a current of 1 A. The coil is placed in a uniform magnetic field of magnitude 0.4 T. The axis of the coil makes an angle of \( 30^\circ \) with the direction of the magnetic field. The torque acting on the coil is \( \alpha \times 10^{-4} \text{ N.m} \). The value of \( \alpha \) is _______. (\( \pi = 3.14 \))

Answer 1

Correct Answer: 314

Step 1: Understanding the Concept:
When a current-carrying coil is placed in a magnetic field, it experiences a magnetic torque. The magnitude of this torque depends on the coil's magnetic moment, the external magnetic field, and the angle between the coil's normal vector (its axis) and the magnetic field.
Step 2: Key Formula or Approach:
Torque on a coil: $\tau = |\vec{M} \times \vec{B}| = M B \sin \theta$
Magnetic moment of the coil: $M = N I A$
Area of circular coil: $A = \pi r^2$
Thus, $\tau = N I A B \sin \theta$.
Step 3: Detailed Explanation:
Identify the given values:
Number of turns, $N = 125$
Current, $I = 1 \text{ A}$
Radius, $r = 2 \text{ cm} = 0.02 \text{ m}$
Magnetic field, $B = 0.4 \text{ T}$
Angle between axis and field, $\theta = 30^\circ$
$\pi = 3.14$
First, calculate the area of the coil $A$:
\[ A = \pi r^2 = 3.14 \times (0.02)^2 = 3.14 \times 0.0004 = 12.56 \times 10^{-4} \text{ m}^2 \] Now, substitute everything into the torque formula:
\[ \tau = N \cdot I \cdot A \cdot B \cdot \sin \theta \] \[ \tau = 125 \times 1 \times (12.56 \times 10^{-4}) \times 0.4 \times \sin(30^\circ) \] Since $\sin(30^\circ) = 0.5$:
\[ \tau = 125 \times 0.4 \times 0.5 \times 12.56 \times 10^{-4} \] \[ \tau = 125 \times 0.2 \times 12.56 \times 10^{-4} \] \[ \tau = 25 \times 12.56 \times 10^{-4} \] \[ \tau = 314 \times 10^{-4} \text{ N.m} \] The problem states the torque is $\alpha \times 10^{-4} \text{ N.m}$.
Matching our result with the required format:
$\alpha = 314$.
Step 4: Final Answer:
The value of $\alpha$ is 314.