Two short bar magnets of length 1 cm each have magnetic moments 1.20 Am^2 and 1.00 Am^2, respectivel

Q: Why is the magnetic field on the equator opposite to the magnetic moment?

This is a fundamental property of magnetic dipoles. If you draw the magnetic field lines originating from the North pole and terminating at the South pole, you will see that at any point on the equatorial plane, the tangent to the field line points exactly opposite to the direction of the dipole moment (which points from South to North).

Q: Why did we take $r = 10$ cm instead of $20$ cm?

The formula for the magnetic field of a dipole uses $r$ as the distance from the *center* of the dipole to the point of observation. The problem states the magnets are separated by $20$ cm, so the midpoint $O$ is exactly half that distance, which is $10$ cm away from each magnet.

Q: How do we know the direction of Earth's horizontal magnetic field?

By convention, the Earth's horizontal magnetic field ($B_H$) always points from the geographic South towards the geographic North. This is the direction a freely suspended compass needle will point.

Q: What if the North poles were pointing North instead of South?

If the North poles pointed North, the magnetic moments would point North. Consequently, the magnetic fields at the equator ($B_1$ and $B_2$) would point South. Since Earth's field $B_H$ points North, you would have to subtract $B_1$ and $B_2$ from $B_H$ to find the net field.

Q: Why did we use the formula for a 'short' bar magnet?

The problem explicitly states 'Two short bar magnets'. When the length of the magnet is much smaller than the distance $r$ at which we are measuring the field, we can ignore the length of the magnet in the denominator, simplifying the exact formula to the 'short dipole' approximation.

Question

Two short bar magnets of length 1 cm each have magnetic moments $1.20 	ext{ Am}^2$ and $1.00 	ext{ Am}^2$, respectively. They are placed on a horizontal table parallel to each other with their N poles pointing towards the South. They have a common magnetic equator and are separated by a distance of 20.0 cm. The value of the resultant horizontal magnetic induction at the mid-point O of the line joining their centres is close to (Horizontal component of the earth's magnetic induction is $3.6 	imes 10^{-5} 	ext{ Wb/m}^2$)

Options:

(A) $3.6 	imes 10^{-5} 	ext{ Wb/m}^2$

(B) $2.56 	imes 10^{-4} 	ext{ Wb/m}^2$

(C) $3.50 	imes 10^{-4} 	ext{ Wb/m}^2$

(D) $5.80 	imes 10^{-4} 	ext{ Wb/m}^2$

Sigma Percentile · Accepted Answer

The beauty of physics often lies in the invisible forces that surround us. In this problem, we are tasked with finding the net magnetic field at a specific point in space, created by a combination of artificial magnets and the Earth itself. It is a classic exercise in vector addition and understanding spatial orientations.

### Decoding the Setup

Imagine you are looking down at a horizontal table. There are two short bar magnets placed parallel to each other. The problem states that their North poles are pointing towards the geographic South. This is a crucial detail!

They share a common magnetic equator, which is an imaginary line perpendicular to their axes, passing through their centers. We need to find the resultant magnetic field at the midpoint $O$ of the line joining their centers. The total distance between the magnets is $20.0 	ext{ cm}$, which means the distance $r$ from the midpoint $O$ to each magnet is exactly $10.0 	ext{ cm}$ or $0.1 	ext{ m}$.

### The Secret of Directions

Before we jump into formulas, we must establish the directions of the magnetic fields. This is where many students make a silly mistake.

The magnetic moment $\vec{M}$ of any magnet is always directed from its South pole to its North pole. Since the North poles of our magnets are pointing South, their magnetic moments are directed Southwards.

Now, recall the behavior of a magnetic field on the equatorial line of a bar magnet. The magnetic field $\vec{B}$ at any point on the equatorial line is always **anti-parallel** (opposite) to the magnetic moment $\vec{M}$.

Since $\vec{M}$ is pointing Southwards, the magnetic fields $\vec{B}_1$ and $\vec{B}_2$ produced by the two magnets at point $O$ must point **Northwards**.

### Enter the Earth

But the magnets aren't the only players in this game. The Earth itself is a giant magnet. The horizontal component of the Earth's magnetic field, denoted as $\vec{B}_H$, naturally points from the geographic South to the geographic North.

Take a moment to realize what this means: $\vec{B}_1$, $\vec{B}_2$, and $\vec{B}_H$ are all pointing in the exact same direction—Northwards! Because they are collinear and point the same way, finding the net magnetic field is as simple as adding their magnitudes together.

$$B_{	ext{net}} = B_1 + B_2 + B_H$$

### The Master Equation

For a short bar magnet, the magnitude of the magnetic field on its equatorial line is given by the formula:

$$B = \frac{\mu_0}{4\pi} \frac{M}{r^3}$$

Let's substitute this into our net field equation for both magnets:

$$B_{	ext{net}} = \frac{\mu_0}{4\pi} \frac{M_1}{r^3} + \frac{\mu_0}{4\pi} \frac{M_2}{r^3} + B_H$$

Since the distance $r$ is the same for both magnets, we can factor out the common terms to make our calculation cleaner:

$$B_{	ext{net}} = \frac{\mu_0}{4\pi} \frac{(M_1 + M_2)}{r^3} + B_H$$

### Crunching the Numbers

Now, we carefully substitute the given values into our master equation. We know that $\frac{\mu_0}{4\pi} = 10^{-7} 	ext{ T m/A}$. The magnetic moments are $M_1 = 1.20 	ext{ Am}^2$ and $M_2 = 1.00 	ext{ Am}^2$. The distance $r = 0.1 	ext{ m}$.

$$B_{	ext{net}} = \frac{10^{-7} (1.20 + 1.00)}{(0.1)^3} + 3.6 	imes 10^{-5}$$

Let's simplify the denominator. $(0.1)^3$ is $10^{-3}$.

$$B_{	ext{net}} = \frac{10^{-7} 	imes 2.20}{10^{-3}} + 3.6 	imes 10^{-5}$$

Dividing $10^{-7}$ by $10^{-3}$ gives us $10^{-4}$.

$$B_{	ext{net}} = 2.20 	imes 10^{-4} + 3.6 	imes 10^{-5}$$

To add these two terms easily, let's express Earth's magnetic field in terms of $10^{-4}$ as well. $3.6 	imes 10^{-5}$ is the same as $0.36 	imes 10^{-4}$.

$$B_{	ext{net}} = 2.20 	imes 10^{-4} + 0.36 	imes 10^{-4}$$

$$B_{	ext{net}} = 2.56 	imes 10^{-4} 	ext{ Wb/m}^2$$

And there we have it! By carefully analyzing the directions of the magnetic vectors and applying the principle of superposition, we arrived at the correct resultant magnetic induction.

Select Answer:

Visualized Solution

Visualizing the Setup

Direction of Magnetic Fields

Earth's Horizontal Magnetic Field

Formula for Equatorial Field

Substituting the Values

Calculating the Net Field

Conclusion

The Sigma Insight: Bar Magnet

Decoding the Setup

The Secret of Directions

Enter the Earth

The Master Equation

Crunching the Numbers

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