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

Question

Two identical short bar magnets, each of magnetic moment $M$, are placed perpendicular to each other at a distance $d$ apart. The magnetic field at a point midway between them on the line joining their centers is:

SUBHANI · Accepted Answer

## CORE INFORMATION
✓ **Answer**: B
## CONCEPT FRAMEWORK
	**Primary Concept**: Magnetic field due to a bar magnet
	**Related Topics**: Magnetic dipole, Vector addition
	**Prerequisites**: Biot-Savart Law, Magnetic moment
## SOLUTION PATHWAY
**Given Information**:
	Two identical short bar magnets with magnetic moment $M$
	Distance between their centers: $d$
	Point of interest: Midway between the magnets
**Method & Steps**:
	1. **Field due to each magnet**:
		Reasoning: The magnetic field at an axial point due to a short bar magnet is given by $B_{axial} = \frac{2\mu_0 M}{4\pi r^3}$, and at an equatorial point it is $B_{equatorial} = \frac{\mu_0 M}{4\pi r^3}$.
		Formula/Rule: $B_{axial} = \frac{2\mu_0 M}{4\pi r^3}$, $B_{equatorial} = \frac{\mu_0 M}{4\pi r^3}$
	2. **Field at the midpoint**:
		Working: Here, $r = d/2$. One magnet is placed axially, and the other equatorially to the midpoint. So, $B_1 = \frac{2\mu_0 M}{4\pi (d/2)^3} = \frac{16\mu_0 M}{4\pi d^3} = \frac{4\mu_0 M}{\pi d^3}$ and $B_2 = \frac{\mu_0 M}{4\pi (d/2)^3} = \frac{8\mu_0 M}{4\pi d^3} = \frac{2\mu_0 M}{\pi d^3}$.
		Key Point: The fields are perpendicular to each other.
	3. **Resultant field**:
		Result: The resultant field is $B = \sqrt{B_1^2 + B_2^2} = \sqrt{(\frac{4\mu_0 M}{\pi d^3})^2 + (\frac{2\mu_0 M}{\pi d^3})^2} = \sqrt{\frac{20\mu_0^2 M^2}{\pi^2 d^6}} = \frac{2\sqrt{5}\mu_0 M}{\pi d^3}$.
		Verification: The answer matches option B.
## OPTION ANALYSIS
[A] :x: Incorrect. This is the result of an incorrect calculation of the magnitude of the resultant field.
[B] ✓: Correct. This is the correct magnitude of the resultant magnetic field at the midpoint.
[C] :x: Incorrect. This likely arises from confusing the axial and equatorial fields or an error in the calculation.
[D] :x: Incorrect. This might be due to doubling the correct resultant field.
## LEARNING AIDS
	**Common Mistakes**:
		1. Incorrectly using the formulas for axial and equatorial fields: Ensure the correct formula is used based on the orientation of the magnet.
		2. Forgetting to take the vector sum: Remember that magnetic fields are vectors and must be added accordingly.
	**Quick Tips**:
		1. Remember the ratio of axial to equatorial field is 2:1.
		2. Visualize the fields as vectors to help determine the direction and magnitude of the resultant field.

SUBHANI · Answer

\frac{\sqrt{5} \mu_0 M}{2\pi d^3}

SUBHANI · Answer

The correct Option is "\frac{2\sqrt{5} \mu_0 M}{\pi d^3}"

SUBHANI · Answer

\frac{2 \mu_0 M}{\sqrt{5}\pi d^3}

SUBHANI · Answer

\frac{4\sqrt{5} \mu_0 M}{\pi d^3}

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