Question 4 of 9beginner🔧 ApplyNumerical2 marks

A short bar magnet placed with its axis at 30° with a uniform external magnetic field of 0.25 T experiences a torque of magnitude equal to 4.5×104.5 \times 10–2 J. What is the magnitude of magnetic moment of the magnet?

Correct Answer

The magnitude of the magnetic moment of the magnet is 0.36 J T⁻¹.

Given:

  • Torque (τ) = 4.5×104.5 \times 10⁻² J
  • Magnetic field (B) = 0.25 T
  • Angle (θ) = 30°

Formula: The torque on a bar magnet placed in a uniform magnetic field is given by: τ = mB sinθ

Calculation: Rearranging the formula to find magnetic moment (m): m = τ / (B sinθ)

Substituting the values: m = (4.5×104.5 \times 10⁻²) / (0.25 × sin 30°) m = (4.5×104.5 \times 10⁻²) / (0.25×0.50.25 \times 0.5) m = (4.5×104.5 \times 10⁻²) / 0.125 m = 0.36 J T⁻¹

Therefore, the magnitude of the magnetic moment of the magnet is 0.36 J T⁻¹.

Exercise: EXERCISES | Q: 5.1 | (Page 152)
For More Understanding

Explanation

This question tests the student's understanding of the relationship between torque and magnetic moment. From the textbook context, the torque on a magnetic dipole in a uniform magnetic field is given by τ = m × B, and in magnitude τ = mB sinθ, where m is the magnetic moment, B is the magnetic field, and θ is the angle between m and B. The student needs to rearrange this formula to solve for m.

Solution Steps

  1. Step 1: Identify the given values - τ = 4.5×104.5 \times 10⁻² J, B = 0.25 T, θ = 30°

  2. Step 2: Write the formula for torque: τ = mB sinθ

  3. Step 3: Rearrange to find m: m = τ / (B sinθ)

  4. Step 4: Substitute values: m = (4.5×104.5 \times 10⁻²) / (0.25 × sin 30°)

  5. Step 5: Calculate: m = (4.5×104.5 \times 10⁻²) / (0.25×0.50.25 \times 0.5) = 0.36 J T⁻¹

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