A magnetic needle of magnetic moment 6.7 x 10^-2 Am^2 and moment of inertia 7.5 x 10^-6 kg m^2 is pe

Q: Why does the needle oscillate instead of just aligning with the field?

When displaced, the magnetic field exerts a restoring torque ($\tau = -MB \sin\theta$). This torque causes angular acceleration back towards equilibrium, leading to Simple Harmonic Motion for small angles.

Q: What happens if the magnetic field is made stronger?

A stronger magnetic field increases the restoring torque. This makes the needle snap back faster, thereby decreasing the time period of oscillation.

Q: Why did we multiply by 10 at the end?

The formula $T = 2\pi \sqrt{I/MB}$ gives the time for exactly *one* complete cycle (back and forth). The question specifically asked for the time taken for 10 complete oscillations.

Q: Is the angle of oscillation important here?

The formula assumes the angle of displacement is small (typically less than 15 degrees) so that $\sin\theta \approx \theta$. For large angles, the motion is no longer perfectly simple harmonic.

Q: How does the mass of the needle affect the time period?

The mass is hidden inside the moment of inertia ($I$). A heavier needle has a larger $I$, making it more sluggish to rotate, which increases the time period.

Question

A magnetic needle of magnetic moment $6.7 	imes 10^{-2} 	ext{ Am}^2$ and moment of inertia $7.5 	imes 10^{-6} 	ext{ kg m}^2$ is performing simple harmonic oscillations in a magnetic field of $0.01 	ext{ T}$. Time taken for 10 complete oscillations is

Options:

(A) $8.89 	ext{ s}$

(B) $6.98 	ext{ s}$

(C) $8.76 	ext{ s}$

(D) $6.65 	ext{ s}$

Sigma Percentile · Accepted Answer

### The Dancing Magnet

Imagine a compass needle resting peacefully, perfectly aligned with the Earth's magnetic field. Now, give it a tiny nudge. What happens? It doesn't just stop; it starts to dance! It swings back and forth, overshooting its resting place again and again.

This rhythmic dance is what physicists call **Simple Harmonic Motion (SHM)**. In our problem, we have a magnetic needle doing exactly this in a uniform magnetic field. The magnetic field acts like an invisible set of springs, constantly trying to pull the needle back to alignment.

### The Master Equation

To understand how fast this needle dances, we need a master equation. The time it takes for the needle to complete one full swing—out and back—is called its **Time Period ($T$)**.

The formula that governs this is a beautiful piece of physics:

$$T = 2\pi \sqrt{\frac{I}{MB}}$$

Let's break down the characters in this equation. $I$ is the **moment of inertia**, which tells us how sluggish the needle is to rotate. A heavier or longer needle has a larger $I$, making the dance slower.

On the flip side, $M$ is the **magnetic moment** (how strong the magnet is), and $B$ is the **external magnetic field**. Together, $MB$ determines the strength of the invisible "springs." A stronger magnet or a stronger field means a faster, snappier dance!

### Crunching the Numbers

Now, let's bring in the numbers from our specific scenario. We are given:
- Moment of inertia, $I = 7.5 	imes 10^{-6} 	ext{ kg m}^2$
- Magnetic moment, $M = 6.7 	imes 10^{-2} 	ext{ Am}^2$
- Magnetic field, $B = 0.01 	ext{ T}$

Let's carefully substitute these into our master equation:

$$T = 2\pi \sqrt{\frac{7.5 	imes 10^{-6}}{6.7 	imes 10^{-2} 	imes 0.01}}$$

First, we simplify the denominator. Multiplying the magnetic moment by the magnetic field:

$$M 	imes B = 6.7 	imes 10^{-2} 	imes 10^{-2} = 6.7 	imes 10^{-4}$$

Now, our equation looks a bit cleaner:

$$T = 2\pi \sqrt{\frac{7.5 	imes 10^{-6}}{6.7 	imes 10^{-4}}}$$

Next, we handle the powers of ten. Dividing $10^{-6}$ by $10^{-4}$ leaves us with $10^{-2}$:

$$T = 2\pi \sqrt{\frac{7.5}{6.7} 	imes 10^{-2}}$$

Taking the square root of $10^{-2}$ gives us $10^{-1}$. The square root of the fraction $7.5 / 6.7$ is approximately $1.058$.

$$T \approx 2\pi 	imes 1.058 	imes 10^{-1}$$

Multiplying this out, we find the time period for a single oscillation:

$$T \approx 0.665 	ext{ s}$$

### The Final Stretch

We have the time for one dance move, but the question asks for a full routine! Specifically, it wants the time taken for **10 complete oscillations**.

This is a classic trap where students might stop too early. We must multiply our single time period by 10:

$$t = 10 	imes T$$

$$t = 10 	imes 0.665 = 6.65 	ext{ s}$$

And there we have it! The needle will take exactly **6.65 seconds** to complete its 10 oscillations.

Select Answer:

Visualized Solution

$Visualizing the Setup$

$Time Period Formula$

$Substituting Values$

$Simplifying the Denominator$

$Handling Powers of Ten$

$Calculating Single Time Period$

$Total Time for 10 Oscillations$

$The Way Forward$

The Sigma Insight: Bar Magnet

The Dancing Magnet

The Master Equation

Crunching the Numbers

The Final Stretch

Similar Questions

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Select Answer:

Visualized Solution

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

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

.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; } Simplifying the Denominator

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Handling Powers of Ten

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Single Time Period

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Total Time for 10 Oscillations

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