At an angle of 30 degrees to the magnetic meridian, the apparent dip is 45 degrees. Find the true di

Q: Why does the vertical component $B_V$ remain the same in the apparent plane?

The apparent plane is just a vertical plane rotated by an angle $\theta$ around the vertical axis. Since the vertical axis itself doesn't change during this rotation, the vertical component of the magnetic field, $B_V$, remains completely unaffected.

Q: Why does the horizontal component become $B_H \cos \theta$?

Think of $B_H$ as a vector lying on the ground. When you look at it from a plane rotated by an angle $\theta$, you are only seeing its projection along that new plane. By basic vector resolution, the projection of $B_H$ along a line at angle $\theta$ is $B_H \cos \theta$.

Q: Can the apparent dip ever be less than the true dip?

No, never! Since $\cos \theta \le 1$, the apparent horizontal component $B_H'$ is always less than or equal to $B_H$. Because $B_H'$ is in the denominator of $\tan \delta' = \frac{B_V}{B_H'}$, a smaller denominator means a larger tangent value, making the apparent dip always greater than or equal to the true dip.

Q: What happens if we measure the dip in a plane perpendicular to the magnetic meridian?

If the plane is perpendicular, $\theta = 90^\circ$. The horizontal component in this plane becomes $B_H \cos 90^\circ = 0$. The dip needle will only feel the vertical force $B_V$, causing it to point straight down, giving an apparent dip of $90^\circ$.

Q: Is the true dip always measured exactly in the magnetic meridian?

Yes, by definition! The true dip is the actual angle of the Earth's magnetic field with the horizontal. To see the full, unprojected magnetic field vector, you must be in the plane that contains both its horizontal and vertical components—which is the magnetic meridian.

Question

At an angle of $30^\circ$ to the magnetic meridian, the apparent dip is $45^\circ$. Find the true dip.

Options:

(A) $	an^{-1}(\sqrt{3})$

(B) $	an^{-1}\left(\frac{1}{\sqrt{3}}ight)$

(C) $	an^{-1}\left(\frac{2}{\sqrt{3}}ight)$

(D) $	an^{-1}\left(\frac{\sqrt{3}}{2}ight)$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in an open field holding a highly sensitive magnetic compass. If you align yourself perfectly with the Earth's magnetic field, you are standing in the **magnetic meridian**. In this plane, the magnetic needle dips downwards at an angle called the **true dip** ($\delta$).

However, what happens if you turn away from the magnetic meridian by an angle $	heta$? You are now in a new vertical plane. The vertical pull of the Earth's magnetic field, $B_V$, remains exactly the same because you haven't tilted the plane up or down. But the horizontal pull, $B_H$, is no longer fully acting in your new plane. You only feel its projection, which is $B_H' = B_H \cos 	heta$. Because the horizontal pull is weaker, the needle dips even steeper! This new, steeper angle is called the **apparent dip** ($\delta'$).

### The Master Equation

To solve this problem, we need to connect the apparent dip to the true dip. In our new plane, the apparent dip is governed by the ratio of the vertical component to the new horizontal component.

$$ 	an \delta' = \frac{B_V}{B_H'} $$

Since we know that $B_H' = B_H \cos 	heta$, we can rewrite this as:

$$ 	an \delta' = \frac{B_V}{B_H \cos 	heta} $$

This is our master equation. It beautifully links the apparent dip ($\delta'$), the true dip components ($B_V$ and $B_H$), and the angle of deviation ($	heta$).

### Substituting the Given Values

The problem states that we are at an angle $	heta = 30^\circ$ to the magnetic meridian, and the apparent dip observed is $\delta' = 45^\circ$. Let's carefully substitute these values into our master equation.

$$ 	an 45^\circ = \frac{B_V}{B_H \cos 30^\circ} $$

We know from basic trigonometry that $	an 45^\circ = 1$ and $\cos 30^\circ = \frac{\sqrt{3}}{2}$. Substituting these standard values gives us:

$$ 1 = \frac{B_V}{B_H \left(\frac{\sqrt{3}}{2}ight)} $$

### Finding the True Dip

Now, let's rearrange this equation to find a direct relationship between $B_V$ and $B_H$. By cross-multiplying, we get:

$$ 1 = \frac{2 B_V}{\sqrt{3} B_H} $$

$$ \Rightarrow B_V = \frac{\sqrt{3}}{2} B_H $$

We are almost there! The question asks for the true dip, $\delta$. By definition, the true dip is the angle in the actual magnetic meridian, given by:

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

Let's substitute the relationship we just found for $B_V$ into this true dip equation:

$$ 	an \delta = \frac{\frac{\sqrt{3}}{2} B_H}{B_H} $$

The $B_H$ terms elegantly cancel out, leaving us with:

$$ 	an \delta = \frac{\sqrt{3}}{2} $$

Finally, taking the inverse tangent gives us the true dip:

$$ \delta = 	an^{-1}\left(\frac{\sqrt{3}}{2}ight) $$

This matches option (d). The beauty of this problem lies in understanding that while the horizontal component of the Earth's magnetic field changes as you rotate your plane of observation, the vertical component remains a steadfast anchor.

Animated Solution for Physics - Magnetism and Matter: At an angle of $3 0^{\circ}$ to the magnetic meridian, the apparent dip is $4 5^{\circ}$ . Find the true dip.

Select Answer:

Visualized Solution

Visualizing True and Apparent Dip

Formula for Apparent Dip

Substituting Given Values

Evaluating Trigonometric Values

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

Formula for True Dip

Substituting $B_{V}$

Final Answer

The Way Forward

The Sigma Insight: Earth Magnetism

Analyzing the Setup

The Master Equation

Substituting the Given Values

Finding the True Dip

Animated Solution for Physics - Magnetism and Matter: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } At an angle of 30∘ to the magnetic meridian, the apparent dip is 45∘. Find the true dip.