Question

At a place where the magnitude of the earth's magnetic field is \(4 \times 10^{-5} \, T\), a short bar magnet is placed with its axis perpendicular to the earth's magnetic field direction. If the resultant magnetic field at a point at a distance of 40 cm from the center of the magnet on the normal bisector of the magnet is inclined at \(45^\circ\) with the earth's field, the magnetic moment of the magnet is:

• A. \(38.4 \, {Am}^2\)
• B. \(51.2 \, {Am}^2\)
• C. \(12.8 \, {Am}^2\)
• D. \(25.6 \, {Am}^2\)

Hint:

Always remember that at the angle of \(45^\circ\), the components of two vector fields are equal, which simplifies the calculation for their magnitudes.

Answer 1

The Correct Option is D

Step 1: Recognize that the magnet's magnetic field and the Earth's magnetic field are combining to form a resultant field at the given point. 
Step 2: Since the resultant field makes a \(45^\circ\) angle with the Earth's magnetic field, the magnitudes of the Earth's field and the magnet's field at that point are equal. Thus: \[ B_{{magnet}} = B_{{earth}} = 4 \times 10^{-5} \, T \] Step 3: Use the formula for the magnetic field due to a dipole at the midpoint of its perpendicular bisector: \[ B = \frac{\mu_0}{4\pi} \frac{2m}{r^3} \] Where: - \( B \) is the magnetic field at the point on the normal bisector, - \( m \) is the magnetic moment of the magnet, - \( r \) is the distance from the magnet to the point (here, \(r = 0.4 \, {m}\)), - \( \mu_0 = 4\pi \times 10^{-7} \, {T} \cdot {m/A} \). 
Step 4: Solve for \(m\) (magnetic moment of the magnet): \[ B = \frac{\mu_0}{4\pi} \frac{2m}{r^3} \] \[ 4 \times 10^{-5} = \frac{4\pi \times 10^{-7}}{4\pi} \frac{2m}{(0.4)^3} \] \[ 4 \times 10^{-5} = 10^{-7} \frac{2m}{0.064} \] \[ 4 \times 10^{-5} \times 0.064 = 2 \times 10^{-7} m \] \[ 2.56 \times 10^{-6} = 2 \times 10^{-7} m \] \[ m = \frac{2.56 \times 10^{-6}}{2 \times 10^{-7}} = 25.6 \, {Am}^2 \]