Question

The magnetic field at a point P on the axis of a short bar magnet of magnetic moment M is B. If another short bar magnet of magnetic moment 2M is placed on the first magnet such that their axes are perpendicular and their centres coincide. The resultant magnetic field at the point P due to both the magnets is

• A. 3B
• B. \(\sqrt{3}\)B
• C. \(\sqrt{2}\)B
• D. 2B

Hint:

Remember the key relationship for a short magnet: \(B_{axis} = 2 \times B_{equatorial}\) for the same magnet at the same distance. In this problem, the second magnet has twice the moment (2M) but P is on its equatorial line, which exactly cancels out the factor of 2 difference, making the two fields equal in magnitude.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
This problem involves the superposition of magnetic fields. We need to find the magnetic field produced by each of the two short bar magnets at point P and then find their vector sum to get the resultant field.
Step 2: Key Formula or Approach:
For a short bar magnet of magnetic moment \(M\), at a distance \(r\) from its center:
- Magnetic field on its axis: \(B_{axis} = \frac{\mu_0}{4\pi} \frac{2M}{r^3}\).
- Magnetic field on its equatorial line: \(B_{equatorial} = \frac{\mu_0}{4\pi} \frac{M}{r^3}\).
The resultant field from two perpendicular vectors \(B_1\) and \(B_2\) is \(B_{res} = \sqrt{B_1^2 + B_2^2}\).
Step 3: Detailed Explanation:
Let the first magnet (moment M) be placed along the x-axis with its center at the origin. Point P is on its axis at a distance \(r\), so P = (r, 0).
The magnetic field at P due to the first magnet is \(B_1\), directed along the x-axis.
\[ B_1 = B_{axis} = \frac{\mu_0}{4\pi} \frac{2M}{r^3} \] We are given that this field has a magnitude \(B\). So, \(B_1 = B\).
The second magnet (moment 2M) is placed with its center at the origin and its axis perpendicular to the first magnet. Let's say its axis is along the y-axis.
For this second magnet, the point P = (r, 0) lies on its equatorial line.
The magnetic field at P due to the second magnet is \(B_2\), directed along the y-axis (or -y, but perpendicular to \(B_1\)).
\[ B_2 = B_{equatorial} = \frac{\mu_0}{4\pi} \frac{(\text{moment})}{r^3} = \frac{\mu_0}{4\pi} \frac{2M}{r^3} \] By comparing the expressions for \(B_1\) and \(B_2\), we can see that their magnitudes are equal:
\[ B_2 = B_1 = B \] The total magnetic field at P, \(\vec{B}_{res}\), is the vector sum of \(\vec{B}_1\) and \(\vec{B}_2\). Since \(\vec{B}_1\) is along the x-axis and \(\vec{B}_2\) is along the y-axis, they are perpendicular.
The magnitude of the resultant field is:
\[ |\vec{B}_{res}| = \sqrt{B_1^2 + B_2^2} = \sqrt{B^2 + B^2} = \sqrt{2B^2} = B\sqrt{2} \] Step 4: Final Answer:
The resultant magnetic field at point P is \(\sqrt{2}B\). Therefore, option (C) is correct.