At some location on earth, the horizontal component of earth's magnetic field is 18 x 10^-6 T. At th

Q: Why doesn't the horizontal magnetic field ($B_H$) create any torque?

Torque is calculated as Force multiplied by the perpendicular distance from the pivot. When the needle is horizontal, the horizontal forces ($mB_H$) pull directly outward along the axis of the needle. Because their line of action passes exactly through the pivot point, the perpendicular distance is zero, resulting in zero torque.

Q: Why is the total magnetic torque $mB_V L$ and not $mB_V L/2$?

There are two poles! The vertical field pulls one pole down (creating a torque of $mB_V \times L/2$) and pushes the other pole up (creating another torque of $mB_V \times L/2$ in the same rotational direction). Adding them together gives $mB_V L/2 + mB_V L/2 = mB_V L$.

Q: I tried to use the length $0.12$ m in my calculation, but it didn't seem to matter. Why?

You discovered a beautiful mathematical cancellation! The magnetic torque depends on the full length $L$, but the counter-torque from your applied force depends on the half-length $L/2$. When you equate them ($F \times L/2 = mB_V L$), the $L$ on both sides cancels out completely. The $0.12$ m was a distractor to test your algebraic confidence.

Q: What if the external force was applied halfway between the center and the end?

If the force was applied at a distance of $L/4$ from the pivot, the torque equation would be $F \times L/4 = mB_V L$. Solving for $F$ would give $F = 4mB_V$. You would need twice as much force because you have half the leverage!

Q: Why did we assume $B_V = B_H$?

This comes directly from the definition of the dip angle ($\theta$). The tangent of the dip angle is the ratio of the vertical to horizontal field: $\tan \theta = B_V / B_H$. Since the problem states the dip angle is $45^\circ$, and $\tan 45^\circ = 1$, it mathematically proves that $B_V$ must equal $B_H$.

Question

At some location on earth, the horizontal component of earth's magnetic field is $18 	imes 10^{-6}$ T. At this location, magnetic needle of length $0.12$ m and pole strength $1.8$ A-m is suspended from its mid point using a thread, it makes $45^\circ$ angle with horizontal in equilibrium. To keep this needle horizontal, the vertical force that should be applied at one of its ends is

Options:

(A) $6.5 	imes 10^{-5}$ N

(B) $3.6 	imes 10^{-5}$ N

(C) $1.3 	imes 10^{-5}$ N

(D) $1.8 	imes 10^{-5}$ N

Sigma Percentile · Accepted Answer

### The Anatomy of Earth's Magnetic Field

Imagine you are holding a compass needle suspended perfectly from its center. You might expect it to rest perfectly parallel to the ground, but it doesn't. Instead, it tilts downwards. Why? Because the Earth's magnetic field isn't just horizontal; it dives into the Earth at an angle known as the **dip angle** ($	heta$).

In this problem, the needle dips at an angle of $45^\circ$. The tangent of this dip angle gives us the direct relationship between the vertical component ($B_V$) and the horizontal component ($B_H$) of the Earth's magnetic field:

$$	an 	heta = \frac{B_V}{B_H}$$

Since $	an 45^\circ = 1$, we immediately discover a crucial piece of the puzzle: the vertical magnetic field is exactly equal in magnitude to the horizontal magnetic field.

$$B_V = B_H = 18 	imes 10^{-6} 	ext{ T}$$

### The Tug of War: Analyzing the Torques

Now, the problem asks us to force this naturally dipping needle to stay perfectly horizontal. To do this, we must understand the invisible forces acting on its poles.

The horizontal field ($B_H$) pulls the North and South poles sideways. Because these forces pass directly through the pivot point (the center of the needle), they create absolutely zero rotational twist. They just stretch the needle slightly.

However, the vertical field ($B_V$) is the troublemaker. It pulls the North pole downwards with a force of $mB_V$ and the South pole upwards with an equal force of $mB_V$. Since these forces act at a distance of $L/2$ from the pivot, they create a combined twisting effect, or **torque** ($	au_B$):

$$	au_B = \left(mB_V 	imes \frac{L}{2}ight) + \left(mB_V 	imes \frac{L}{2}ight) = mB_V L$$

### The Illusion of Length: A Beautiful Cancellation

To stop the needle from spinning and keep it horizontal, we must apply an external vertical force $F$ at one of its ends. This applied force will create a counter-torque ($	au_F$) to balance the magnetic torque:

$$	au_F = F 	imes \frac{L}{2}$$

For the needle to be in perfect equilibrium, our applied torque must exactly equal the magnetic torque:

$$F 	imes \frac{L}{2} = mB_V L$$

Here is where the magic of physics happens. Notice how the length of the needle, $L$, appears on both sides of the equation? When we solve for $F$, the length completely cancels out!

$$F = 2 m B_V$$

The exact length of the needle ($0.12 	ext{ m}$) provided in the question is actually a clever distractor. Whether the needle is $10 	ext{ cm}$ or $10 	ext{ m}$ long, the required force at the end remains exactly the same!

### The Final Calculation

With our elegant, simplified formula, all that remains is to plug in the known values. We know the pole strength $m = 1.8 	ext{ A-m}$ and $B_V = 18 	imes 10^{-6} 	ext{ T}$.

$$F = 2 	imes 1.8 	imes 18 	imes 10^{-6}$$

$$F = 3.6 	imes 18 	imes 10^{-6}$$

$$F = 64.8 	imes 10^{-6} 	ext{ N}$$

Adjusting the decimal for standard scientific notation, we arrive at our final answer:

$$F = 6.5 	imes 10^{-5} 	ext{ N}$$

This is the precise force required to conquer the Earth's magnetic pull and keep the needle perfectly flat.

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Visualized Solution

$Understanding the Setup$

$Relation between B_{V} and B_{H}$

$Forces in Horizontal Position$

$Balancing the Torque$

$Equating Torques$

$Substituting Values$

$Final Calculation$

The Sigma Insight: Bar Magnet

The Anatomy of Earth's Magnetic Field

The Tug of War

The Illusion of Length

The Final Calculation

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Understanding the Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Relation between BV​ and BH​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Forces in Horizontal Position

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Balancing the Torque

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

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

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