In a uniform magnetic field, the magnetic needle has a magnetic moment 9.85 x 10^-2 A/m^2 and moment

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

When displaced, the magnetic field exerts a restoring torque ($\tau = -MB \sin\theta$) that pulls it back towards equilibrium. Due to its moment of inertia, it overshoots the equilibrium position, leading to Simple Harmonic Motion.

Q: How did we get the time period $T = 0.5 \text{ s}$?

The time period is the time taken for exactly one complete oscillation. Since it completes 10 oscillations in 5 seconds, one oscillation takes $5 / 10 = 0.5$ seconds.

Q: Why did we square the time period equation?

The unknown variable, the magnetic field $B$, was trapped inside a square root. Squaring both sides is the cleanest algebraic step to isolate $B$ without dealing with messy radical terms.

Q: What is the significance of the hint $\pi^2 = 9.85$?

In competitive exams, examiners often design numbers to cancel out perfectly. Here, the magnetic moment $M$ was given as $9.85 \times 10^{-2}$, which exactly cancels the $\pi^2$ term in the numerator, saving you from tedious decimal calculations.

Q: Why is the final answer 8 and not $8 \times 10^{-3}$?

Always read the units requested in the question carefully! The question specifically asked for the magnitude in milli-Tesla (mT). Since $1 \text{ mT} = 10^{-3} \text{ T}$, $8 \times 10^{-3} \text{ T}$ is exactly $8 \text{ mT}$.

Question

In a uniform magnetic field, the magnetic needle has a magnetic moment $9.85 	imes 10^{-2} 	ext{ A/m}^2$ and moment of inertia $5 	imes 10^{-6} 	ext{ kg-m}^2$. If it performs 10 complete oscillations in 5 s, then the magnitude of the magnetic field is ......... mT. [Take, $\pi^2$ as 9.85]

Sigma Percentile · Accepted Answer

### Setting the Stage: The Oscillating Needle

Imagine a compass needle resting peacefully in a uniform magnetic field. If you give it a tiny nudge, it doesn't just spin out of control.

It experiences a **restoring torque** that pulls it back, causing it to oscillate back and forth, much like a pendulum.

This restoring torque arises because the magnetic field tries to align the magnetic dipole moment of the needle with itself.

### The Master Equation: Mechanics meets Electromagnetism

Because this restoring torque is proportional to the angular displacement for small angles, the needle performs **Simple Harmonic Motion (SHM)**.

The time period of this oscillation is given by a classic formula:

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

This equation beautifully connects the mechanics of inertia ($I$) with the electromagnetism of the dipole ($M$ and $B$).

### Decoding the Given Data

Now, the problem tells us that the needle completes 10 full oscillations in exactly 5 seconds.

The time period is simply the time taken for one single oscillation. So, we divide the total time by the number of oscillations:

$$T = \frac{5}{10} = 0.5 	ext{ s}$$

This gives us a clean time period of $0.5 	ext{ s}$. Simple and sweet!

### The Art of Algebraic Rearrangement

We need to find the magnitude of the magnetic field, $B$. To get $B$ out of that square root prison, let's square both sides of our time period equation.

$$T^2 = 4\pi^2 \left(\frac{I}{MB}ight)$$

Now, just a quick cross-multiplication to isolate $B$ on one side:

$$B = \frac{4\pi^2 I}{M T^2}$$

Watch out for the algebra here; isolating variables correctly is crucial before plugging in any messy numbers.

### The Magic of Cancellation and Final Computation

It's time to plug in the numbers. We have the moment of inertia $I = 5 	imes 10^{-6} 	ext{ kg m}^2$, the magnetic moment $M = 9.85 	imes 10^{-2} 	ext{ A/m}^2$, and our newly found time period $T = 0.5 	ext{ s}$.

Also, notice the beautiful hint given in the question: take $\pi^2$ as $9.85$. Let's substitute all these values into our rearranged equation:

$$B = \frac{4 	imes 9.85 	imes 5 	imes 10^{-6}}{9.85 	imes 10^{-2} 	imes (0.5)^2}$$

Did you get the feel of it? The $9.85$ in the numerator and denominator cancel out perfectly! This is a classic JEE move to test your presence of mind.

$$B = \frac{20 	imes 10^{-6}}{0.25 	imes 10^{-2}}$$

Now, $4 	imes 5$ gives $20$ in the numerator, and $0.5^2$ is $0.25$ in the denominator. The calculation just became incredibly easy.

$$B = \frac{20}{0.25} 	imes 10^{-4} = 80 	imes 10^{-4} 	ext{ T}$$

Which simplifies to exactly $8 	imes 10^{-3} 	ext{ T}$. Since the question asks for the answer in milli-Tesla (mT), our final answer is simply **8**.

Enter Numerical Value:

Visualized Solution

The Oscillating Magnetic Needle

Time Period of Oscillation $T$

Calculating Time Period $T$

Rearranging for Magnetic Field $B$

Substituting the Values

The Magic of Cancellation

Final Calculation for $B$

The Way Forward

The Sigma Insight: Bar Magnet

Setting the Stage

The Master Equation

Decoding the Given Data

The Art of Algebraic Rearrangement

The Magic of Cancellation and Final Computation

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