The Question is from neet-physics magnetic-effects-current-and-magnetism , and it was created by the author P.NARASIMHA

Question

Two identical short bar magnets, each having magnetic moment $M$, are placed a distance of $2d$ apart with axes perpendicular to each other in a horizontal plane. The magnetic induction at a point midway between them is

P.NARASIMHA · Accepted Answer

## CORE INFORMATION
✓ **Answer**: D
## CONCEPT FRAMEWORK
**Primary Concept**: Magnetic field due to a bar magnet and superposition principle.
**Related Topics**: Axial and equatorial magnetic field of a bar magnet, vector addition of magnetic fields.
**Prerequisites**: Understanding of magnetic field expressions for bar magnets, vector addition.
## SOLUTION PATHWAY
**Given Information**:
   - Two identical bar magnets with magnetic moment $M$.
   - Distance between magnets is $2d$.
   - Axes are perpendicular.
   - Point is midway between the magnets.
**Method & Steps**:
   1. **Determine the distance from each magnet to the midpoint:** The midpoint is at a distance $d$ from each magnet.
        Reasoning: The magnets are $2d$ apart, so the midpoint is at half the distance.
        Formula/Rule: Midpoint divides the distance into two equal parts.
   2. **Calculate the magnetic field due to the first magnet:**  Since the point is on the axial line of the first magnet, its magnetic field is given by $B_1 = \frac{{\mu_0}}{{4\pi}} \frac{{2M}}{{{d^3}}}$.
        Working: Axial field formula is applied.
        Key Point: Axial field is twice the equatorial field.
   3. **Calculate the magnetic field due to the second magnet:** Since the point is on the equatorial line of the second magnet, its magnetic field is given by $B_2 = \frac{{\mu_0}}{{4\pi}} \frac{M}{{{d^3}}}$.
        Working: Equatorial field formula is applied.
        Key Point: Equatorial field is half the axial field.
   4. **Apply the superposition principle:** Since the fields are perpendicular, find the resultant magnetic field using vector addition $B_{net} = \sqrt{B_1^2 + B_2^2}$.
        Reasoning: Perpendicular fields are added vectorially.
        Formula/Rule: Pythagorean theorem for vector addition.
   5. **Calculate the net magnetic field:** $B_{net} = \sqrt{\left(\frac{{\mu_0}}{{4\pi}} \frac{{2M}}{{{d^3}}}ight)^2 + \left(\frac{{\mu_0}}{{4\pi}} \frac{M}{{{d^3}}}ight)^2} = \frac{{\mu_0}}{{4\pi}} \frac{M}{{{d^3}}} \sqrt{2^2 + 1^2} = \frac{{\mu_0}}{{4\pi}} \frac{M}{{{d^3}}} \sqrt{5}$
        Result: Resultant field magnitude.
        Verification: Checking the calculation.
 ## OPTION ANALYSIS
[A] :x: Incorrect. This would be the case if both fields were of same magnitude.
[B] :x: Incorrect. No such relation between fields.
[C] :x: Incorrect. Value incorrect.
[D] ✓: Correct. This is the result obtained after correct calculation and vector addition.
 ## LEARNING AIDS
**Common Mistakes**:
   1. **Incorrect use of formulas**: Using axial field formula for equatorial field and vice-versa.
        How to avoid: Clearly identify the position of the point relative to the magnet axis.
   2. **Incorrect vector addition**: Not considering the perpendicular nature of the fields.
        How to avoid: Apply the Pythagorean theorem for vector addition of perpendicular vectors.
**Quick Tips**:
   1. **Remember axial and equatorial field formulas**: Axial field is twice the equatorial field.
   2. **Draw a diagram**: Visualize the problem to understand the direction of magnetic fields.

P.NARASIMHA · Answer

$\frac{{{\mu _0}}}{{4\pi }}\left( {\sqrt 2 } \right)\frac{M}{{{d^3}}}$

P.NARASIMHA · Answer

$\frac{{{\mu _0}}}{{4\pi }}\left( {\sqrt 3 } \right)\frac{M}{{{d^3}}}$

P.NARASIMHA · Answer

$\left( {\frac{{2{\mu _0}}}{\pi }} \right)\frac{M}{{{d^3}}}$

P.NARASIMHA · Answer

The correct Option is "$\frac{{{\mu _0}}}{{4\pi }}\left( {\sqrt 5 } \right)\frac{M}{{{d^3}}}$"

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