Question

At a place the horizontal component of earth’s magnetic field is \( 3 \times 10^{-5} \) T and the magnetic declination is \( 30^\circ \). A compass needle of magnetic moment \( 18 \) A\(m^2\) pointing towards geographic north at this place experiences a torque of:

• A. \( 36 \times 10^{-5} \) Nm
• B. \( 18 \times 10^{-5} \) Nm
• C. \( 54 \times 10^{-5} \) Nm
• D. \( 27 \times 10^{-5} \) Nm

Hint:

The torque on a magnetic dipole in a magnetic field is given by \( \tau = M B \sin\theta \), where \( M \) is the magnetic moment, \( B \) is the magnetic field, and \( \theta \) is the angle between them.

Answer 1

The Correct Option is D

Step 1: Apply Torque Formula The torque experienced by a magnetic dipole in a magnetic field is given by: \[ \tau = M B \sin\theta \] Step 2: Compute Torque \[ \tau = (18) \times (3 \times 10^{-5}) \times \sin 30^\circ \] Since \( \sin 30^\circ = \frac{1}{2} \), we get: \[ \tau = 18 \times 3 \times 10^{-5} \times \frac{1}{2} \] \[ \tau = \frac{54 \times 10^{-5}}{2} \] \[ \tau = 27 \times 10^{-5} \text{ Nm} \] Thus, the correct answer is \( 27 \times 10^{-5} \) Nm.

Answer 2

Given:
- Horizontal component of Earth's magnetic field, \( B_H = 3 \times 10^{-5} \, \text{T} \)
- Magnetic declination, \( \theta = 30^\circ \)
- Magnetic moment of compass needle, \( \mu = 18 \, \text{A} \cdot \text{m}^2 \)

We need to calculate the torque \( \tau \) experienced by the compass needle.

Step 1: Understand the magnetic field components:
- Magnetic declination \( \theta \) is the angle between geographic north and magnetic north.
- The torque on the magnetic needle depends on the component of the Earth's magnetic field perpendicular to the needle's direction.

Step 2: Expression for torque on magnetic dipole in a magnetic field:
\[ \tau = \mu B \sin \phi \] where \( \phi \) is the angle between the magnetic moment vector and the magnetic field.

Step 3: Since the needle is pointing towards geographic north, but the magnetic field is inclined at \( 30^\circ \), the angle between magnetic moment and Earth's field is \( \phi = 30^\circ \).

Step 4: Substitute the values:
\[ \tau = 18 \times 3 \times 10^{-5} \times \sin 30^\circ = 54 \times 10^{-5} \times \frac{1}{2} = 27 \times 10^{-5} \, \text{Nm} \]

Therefore, the torque experienced by the compass needle is:
\[ \boxed{27 \times 10^{-5} \, \text{Nm}} \]