Question

Two identical short bar magnets, each having a magnetic moment of 10 \( \text{Am}^{2} \) are arranged such that their axial lines are perpendicular to each other and their centres be along the same straight line in a horizontal plane. If the distance between their centres is 0.2 m, the resultant magnetic induction at a point midway between them is: (\( \mu_{\circ}=4\pi\times10^{-7}~\text{Hm}^{-1} \))

• A. \( \sqrt{2}\times10^{-7}~\text{T} \)
• B. \( \sqrt{5}\times10^{-7}~\text{T} \)
• C. \( \sqrt{2}\times10^{-3}~\text{T} \)
• D. \( \sqrt{5}\times10^{-3}~\text{T} \)

Hint:

For perpendicular magnetic configurations with a shared distance, the resultant combination always simplifies to a factor of \( \sqrt{2^2 + 1^2} = \sqrt{5} \) times the base equatorial field value, saving you from repeating long scientific calculations.

Answer 1

The Correct Option is D

Step 1: Set up the geometry.
Two identical magnets sit on the same line with their centres $0.2$ m apart. The midpoint is halfway, so each magnet is a distance $d = 0.1$ m from it. Their axes are perpendicular to each other.
Step 2: Recall the two field formulas.
For a short magnet of moment $M$ at distance $d$, the field on its axis is $\dfrac{\mu_0}{4\pi}\dfrac{2M}{d^{3}}$ and on its equator is $\dfrac{\mu_0}{4\pi}\dfrac{M}{d^{3}}$. The midpoint is on the axis of one magnet and on the equator of the other.
Step 3: Compute the base field.
Let $B_0 = \dfrac{\mu_0}{4\pi}\dfrac{M}{d^{3}}$. With $\dfrac{\mu_0}{4\pi} = 10^{-7}$, $M = 10$, $d = 0.1$: \[ B_0 = 10^{-7}\times\frac{10}{(0.1)^{3}} = 10^{-7}\times10^{4} = 10^{-3}\ \text{T} \]
Step 4: Write the two fields.
The axial field is $B_1 = 2B_0$ and the equatorial field is $B_2 = B_0$.
Step 5: Add the perpendicular fields.
Since the two fields are at right angles, combine them with the Pythagoras rule: \[ B_{net} = \sqrt{(2B_0)^{2} + B_0^{2}} = \sqrt{5}\,B_0 \]
Step 6: Put the value in.
\[ B_{net} = \sqrt{5}\times10^{-3}\ \text{T} \] \[ \boxed{B_{net} = \sqrt{5}\times10^{-3}\ \text{T}} \]