A magnetic compass needle oscillates 30 times per minute at a place, where the dip is 45° and 40 tim

Q: Why do we use $B_H$ instead of the total magnetic field $B$ in the time period formula?

A standard magnetic compass is designed with a pivot that restricts its motion strictly to the horizontal plane. Therefore, it can only align with and oscillate around the horizontal component of the Earth's magnetic field ($B_H$). The vertical component ($B_V$) just pulls the needle downwards against the pivot, but doesn't cause horizontal rotation.

Q: Why do the options have the unit 'T' (Tesla) if we are calculating a ratio?

You have a sharp eye! The ratio of two quantities with the same units (like two magnetic fields) is always dimensionless. The inclusion of 'T' in the options is a typographical error in the original JEE question paper. We must ignore the unit and select the option with the correct numerical value.

Q: How did we get $\sqrt{6}$ in the numerator during the calculation?

When we divided $B_{H2}$ by $B_{H1}$, we had a complex fraction: $(\frac{\sqrt{3}}{2}) / (\frac{1}{\sqrt{2}})$. When you divide by a fraction, you multiply by its reciprocal. So, it becomes $\frac{\sqrt{3}}{2} \times \frac{\sqrt{2}}{1}$. Multiplying the numerators gives $\sqrt{3} \times \sqrt{2} = \sqrt{6}$.

Q: What is the difference between a magnetic compass and a dip circle?

A magnetic compass has a needle constrained to rotate horizontally, measuring the direction of $B_H$. A dip circle, on the other hand, has a needle constrained to rotate vertically, allowing it to align with the total magnetic field $B$ and measure the angle of dip.

Q: How can I quickly estimate $\sqrt{6}$ without a calculator during an exam?

Find the nearest perfect squares. $\sqrt{4} = 2$ and $\sqrt{9} = 3$. So $\sqrt{6}$ is between 2 and 3. Since 6 is slightly closer to 4 than 9, the root is slightly less than 2.5. You can quickly check $2.4^2 = 5.76$ and $2.5^2 = 6.25$. So, $2.45$ is a very safe and accurate estimate for exam purposes.

Question

A magnetic compass needle oscillates 30 times per minute at a place, where the dip is $45^{\circ}$ and 40 times per minute, where the dip is $30^{\circ}$. If $B_1$ and $B_2$ are respectively, the total magnetic field due to the earth at the two places, then the ratio $B_1 / B_2$ is best given by

Options:

(A) 1.8 T

(B) 0.7 T

(C) 3.6 T

(D) 2.2 T

Sigma Percentile · Accepted Answer

### Dancing with the Earth's Magnetic Field

Imagine you are holding a standard magnetic compass. When you gently tap the needle, it begins to oscillate back and forth. But what invisible hand is pulling it back to its equilibrium position? It is the Earth's magnetic field. However, the Earth's magnetic field is not perfectly parallel to the ground; it dips downwards at an angle known as the **angle of dip** ($	heta$).

Because a standard compass needle is constrained to rotate only in the horizontal plane, it doesn't feel the full force of the Earth's magnetic field ($B$). Instead, it only responds to the horizontal component, denoted as $B_H$. Using basic trigonometry, we can express this horizontal component as:

$$B_H = B \cos	heta$$

### The Physics of Oscillation

When a magnetic dipole (like our compass needle) with a magnetic moment $M$ and moment of inertia $I$ is displaced in a uniform magnetic field, it experiences a restoring torque. This results in simple harmonic motion. The time period ($T$) of this oscillation is given by the master equation:

$$T = 2\pi \sqrt{\frac{I}{M B_H}}$$

In our specific problem, we are taking the exact same compass needle to two different locations on Earth. This means the physical properties of the needle—its moment of inertia ($I$) and its magnetic moment ($M$)—remain absolutely constant. Therefore, we can establish a beautiful inverse proportionality:

$$T \propto \frac{1}{\sqrt{B_H}}$$

### Analyzing the Two Locations

Let's break down the data given for the two locations. We are given the number of oscillations per minute, which is a measure of frequency. We need to convert this into the time period.

**At the First Location:**
The needle completes 30 oscillations in 60 seconds. 
Frequency, $f_1 = \frac{30}{60} = \frac{1}{2}$ Hz.
The time period is the reciprocal of frequency, so $T_1 = 2$ seconds.
The angle of dip is $	heta_1 = 45^\circ$. Therefore, the horizontal field is:

$$B_{H1} = B_1 \cos 45^\circ = \frac{B_1}{\sqrt{2}}$$

**At the Second Location:**
The needle is faster, completing 40 oscillations in 60 seconds.
Frequency, $f_2 = \frac{40}{60} = \frac{2}{3}$ Hz.
The time period is $T_2 = \frac{3}{2}$ seconds.
The angle of dip is shallower, $	heta_2 = 30^\circ$. Therefore, the horizontal field is:

$$B_{H2} = B_2 \cos 30^\circ = \frac{\sqrt{3}}{2} B_2$$

### The Master Calculation

Now, we bring our proportionality to life by setting up a ratio between the two locations:

$$\frac{T_1}{T_2} = \sqrt{\frac{B_{H2}}{B_{H1}}}$$

Substituting the values we meticulously gathered:

$$\frac{2}{3/2} = \sqrt{\frac{\frac{\sqrt{3}}{2} B_2}{\frac{B_1}{\sqrt{2}}}}$$

Let's simplify the left side to $\frac{4}{3}$. To liberate the magnetic fields from the square root, we must square both sides of the equation:

$$\left(\frac{4}{3}ight)^2 = \frac{\frac{\sqrt{3}}{2} B_2}{\frac{B_1}{\sqrt{2}}}$$

$$\frac{16}{9} = \frac{\sqrt{3} \cdot \sqrt{2}}{2} \frac{B_2}{B_1}$$

$$\frac{16}{9} = \frac{\sqrt{6}}{2} \frac{B_2}{B_1}$$

We are now at the final stretch. We need to isolate the ratio $\frac{B_1}{B_2}$:

$$\frac{B_1}{B_2} = \frac{9 \sqrt{6}}{32}$$

To find the numerical value, we can estimate $\sqrt{6}$. Since $2.4^2 = 5.76$ and $2.5^2 = 6.25$, $\sqrt{6}$ is approximately $2.45$.

$$\frac{B_1}{B_2} \approx \frac{9 	imes 2.45}{32} = \frac{22.05}{32} \approx 0.689$$

Rounding this to one decimal place, we get **0.7**.

*A quick note on units:* You might have noticed that the options in the question include the unit 'T' (Tesla). The ratio of two magnetic fields should be a dimensionless number. This is a typographical error in the original exam paper, but our calculated numerical value confidently points us to option (b).

Select Answer:

Visualized Solution

Earth's Magnetic Field Components

Time Period of Oscillation

Data at First Location

Data at Second Location

Setting up the Ratio

Squaring and Simplifying

Final Calculation

Conclusion

The Way Forward

The Sigma Insight: Bar Magnet

Dancing with the Earth's Magnetic Field

The Physics of Oscillation

Analyzing the Two Locations

The Master Calculation

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