A bar magnet of length 14 cm is placed in the magnetic meridian with its North pole pointing towards

Q: Why does the neutral point lie on the equatorial line?

When the magnet's North pole points North, its magnetic field lines emerge from the North pole and curve around to the South pole. On the equatorial line, these field lines point exactly South, directly opposing the Earth's magnetic field which points North.

Q: Why can't we use the short magnet formula $B = \frac{\mu_0}{4\pi} \frac{M}{r^3}$?

The short magnet approximation is only valid when the distance $r$ is much, much greater than the magnet's half-length $l$. Here, $r = 18$ cm and $l = 7$ cm. Since $l$ is a significant fraction of $r$, the approximation would lead to a highly inaccurate result.

Q: What is the difference between $l$ and $2l$ in the formula?

By convention, the total length of the bar magnet is taken as $2l$ so that the distance from the center to either pole is simply $l$. In this problem, the total length is 14 cm, making $l = 7$ cm.

Q: How do we calculate $(373)^{3/2}$ without a calculator?

You can break it down as $373 \times \sqrt{373}$. To estimate $\sqrt{373}$, find the nearest perfect squares: $19^2 = 361$ and $20^2 = 400$. Since 373 is closer to 361, the square root is roughly 19.3. Multiplying $373 \times 19.3$ gives a very close approximation.

Q: What happens if the magnet's South pole points North?

If the orientation is reversed, the magnet's field on the equatorial line would point North, adding to the Earth's field. The neutral points would then shift to the axial line, where the magnet's field points South and can cancel the Earth's field.

Question

A bar magnet of length 14 cm is placed in the magnetic meridian with its North pole pointing towards the geographic North pole. A neutral point is obtained at a distance of 18 cm from the centre of the magnet. If $B_H = 0.4 	ext{ G}$, then the magnetic moment of the magnet is ($1 	ext{ G} = 10^{-4} 	ext{ T}$)

Options:

(A) $2.88 	imes 10^3 	ext{ J T}^{-1}$

(B) $2.88 	imes 10^2 	ext{ J T}^{-1}$

(C) $2.88 	ext{ J T}^{-1}$

(D) $28.8 	ext{ J T}^{-1}$

Sigma Percentile · Accepted Answer

The problem of finding the magnetic moment of a bar magnet placed in the Earth's magnetic field is a classic test of both conceptual clarity and algebraic stamina. Let's break down the physics and the math behind this fascinating setup.

### The Setup: Visualizing the Magnetic Battlefield

Imagine a bar magnet resting on a table, with its North pole pointing exactly towards the Earth's geographic North. The Earth's own magnetic field, $B_H$, is constantly flowing from South to North.

But our magnet is also creating its own field. On the equatorial line (the line perpendicular to the center of the magnet), the magnet's field points from North to South, directly opposing the Earth's field. The exact spot where these two invisible forces perfectly cancel each other out is called the **neutral point**.

In our problem, this neutral point is located at a distance of $r = 18 	ext{ cm}$ from the center of the magnet.

### The Master Equation: Balancing the Fields

At the neutral point, the net magnetic field is zero. This means the magnetic field produced by the magnet on its equatorial line ($B_{eq}$) must perfectly balance the Earth's horizontal magnetic field ($B_H$).

$$B_{eq} = B_H$$

To find $B_{eq}$, we look at the magnetic field vectors from the North and South poles. The North pole pushes the magnetic field away, while the South pole pulls it in. By resolving these vectors, the horizontal components cancel out, and the vertical components add up to give the net equatorial field:

$$B_{eq} = 2 B_0 \cos	heta$$

Here, $B_0$ is the magnetic field due to a single pole:

$$B_0 = \frac{\mu_0 m}{4\pi (r^2 + l^2)}$$

From the geometry of the setup, $\cos	heta$ is the adjacent side $l$ divided by the hypotenuse $\sqrt{r^2 + l^2}$. Substituting these into our balance equation gives us the complete expression:

$$B_H = 2 \left( \frac{\mu_0 m}{4\pi (r^2 + l^2)} ight) \frac{l}{\sqrt{r^2 + l^2}}$$

Notice the term $m 	imes 2l$ in the numerator. This is exactly the definition of the **magnetic dipole moment**, $M$! Substituting $M$, our equation simplifies beautifully:

$$B_H = \frac{\mu_0}{4\pi} \frac{M}{(r^2 + l^2)^{3/2}}$$

### The Calculation: Taming the Numbers

Our goal is to find the magnetic moment, $M$. Let's rearrange the equation to isolate $M$:

$$M = \frac{B_H \cdot (r^2 + l^2)^{3/2}}{\frac{\mu_0}{4\pi}}$$

Now, we carefully substitute the given values, ensuring all units are in standard SI format:
- $B_H = 0.4 	ext{ G} = 0.4 	imes 10^{-4} 	ext{ T}$
- $r = 18 	ext{ cm} = 0.18 	ext{ m}$
- $l = 7 	ext{ cm} = 0.07 	ext{ m}$
- $\frac{\mu_0}{4\pi} = 10^{-7} 	ext{ T m A}^{-1}$

Plugging these in:

$$M = \frac{0.4 	imes 10^{-4} 	imes (0.18^2 + 0.07^2)^{3/2}}{10^{-7}}$$

Let's tackle the calculation inside the bracket first. 
$0.18^2 = 0.0324$ and $0.07^2 = 0.0049$. 
Adding them up gives $0.0373$, which we can write as $373 	imes 10^{-4}$ to make the square root easier later.

$$M = \frac{0.4 	imes 10^{-4} 	imes (373 	imes 10^{-4})^{3/2}}{10^{-7}}$$

Simplifying the powers of ten: $(10^{-4})^{3/2}$ becomes $10^{-6}$. Bringing the $10^{-7}$ up from the denominator changes its sign to $10^7$. Combining all the powers of ten simplifies our expression significantly:

$$M = 0.4 	imes 10^{-3} 	imes 373^{3/2}$$

### The Final Result

Here is where many students get stuck: calculating $373^{3/2}$. Don't get intimidated! It's just $373 	imes \sqrt{373}$.

Since $19^2 = 361$, the square root of 373 is slightly more than 19, roughly $19.31$. Multiplying these gives approximately $7203.8$.

$$M = 0.4 	imes 10^{-3} 	imes 7203.8$$

This gives us $2881.5 	imes 10^{-3}$. Shifting the decimal point, we get our final answer:

$$M \approx 2.88 	ext{ J T}^{-1}$$

A beautiful result after a rigorous calculation! Always remember to double-check your unit conversions and take your time with the algebraic simplifications.

Select Answer:

Visualized Solution

The Setup

Magnetic Field at Neutral Point

Formulating the Equation

Introducing Magnetic Moment

Rearranging for $M$

Substituting Values

Calculating the Bracket

Simplifying Powers

Evaluating $37 3^{3/2}$

Final Magnetic Moment

What if the magnet is reversed?

The Sigma Insight: Bar Magnet and Magnetic Dipole

The Setup

The Master Equation

The Calculation

The Final Result

Similar Questions

A magnetic needle is kept in a non-uniform magnetic field. It experiences

Select Answer:

Visualized Solution

The Setup

Magnetic Field at Neutral Point

Formulating the Equation

Introducing Magnetic Moment

Rearranging for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } M

Substituting Values

Calculating the Bracket

Simplifying Powers

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 3733/2

Final Magnetic Moment

What if the magnet is reversed?

The Sigma Insight: Bar Magnet and Magnetic Dipole

The Setup

The Master Equation

The Calculation

The Final Result

Similar Questions

A magnetic needle is kept in a non-uniform magnetic field. It experiences

Rearranging for $M$

Evaluating $37 3^{3/2}$