Question

A magnetic needle free to rotate in a vertical plane parallel to the magnetic meridian has its north tip pointing down at \(30^\circ\) with the horizontal. The horizontal component of the earth's magnetic field at the place is \(0.3\;G\). Then the magnitude of the earth's magnetic field at the location is

• A. \(\dfrac{\sqrt{3}}{5}\;G\)
• B. \(\sqrt{3}\;G\)
• C. \(\dfrac{20}{\sqrt{3}}\;G\)
• D. \(\dfrac{2}{\sqrt{3}}\;G\)

Hint:

For earth's magnetic field, \[ H=B\cos\delta \] where \(H\) is the horizontal component, \(B\) is the total field, and \(\delta\) is the angle of dip.

Answer 1

The Correct Option is A

Step 1: Identify the angle of dip.
The needle swings in the vertical magnetic meridian and its north tip dips $30^\circ$ below the horizontal, so the angle of dip is \[ \delta = 30^\circ \]
Step 2: Recall the horizontal component relation.
The horizontal component $H$ relates to the total field $B$ by \[ H = B\cos\delta \]
Step 3: Rearrange for the total field.
Solving for $B$, \[ B = \frac{H}{\cos\delta} \]
Step 4: Insert the known values.
With $H = 0.3\ \text{G}$ and $\cos 30^\circ = \dfrac{\sqrt{3}}{2}$, \[ B = \frac{0.3}{\frac{\sqrt{3}}{2}} = \frac{0.6}{\sqrt{3}} \]
Step 5: Simplify the surd.
Rationalising, \[ B = \frac{0.6}{\sqrt{3}} = 0.2\sqrt{3} = \frac{\sqrt{3}}{5}\ \text{G} \]
Step 6: State the result.
Hence the magnitude of the earth's magnetic field is \[ \boxed{\dfrac{\sqrt{3}}{5}\ \text{G}} \]