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 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
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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.
Updated On: Jun 22, 2026
- \(\dfrac{\sqrt{3}}{5}\;G\)
- \(\sqrt{3}\;G\)
- \(\dfrac{20}{\sqrt{3}}\;G\)
- \(\dfrac{2}{\sqrt{3}}\;G\)
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The Correct Option is A
Solution and Explanation
Step 1: Identify the angle of dip.
The magnetic needle is free to rotate in a vertical plane parallel to the magnetic meridian.
Its north tip points downward at \(30^\circ\) with the horizontal.
Therefore, the angle of dip is \[ \delta=30^\circ \]
Step 2: Use the relation between total magnetic field and horizontal component.
If \(B\) is the magnitude of earth's magnetic field and \(H\) is its horizontal component, then \[ H=B\cos\delta \] Given, \[ H=0.3\;G \] and \[ \delta=30^\circ \]
Step 3: Substitute the values.
\[ 0.3=B\cos30^\circ \] \[ 0.3=B\left(\frac{\sqrt{3}}{2}\right) \] \[ B=\frac{0.3\times2}{\sqrt{3}} \] \[ B=\frac{0.6}{\sqrt{3}} \] \[ B=\frac{3}{5\sqrt{3}} \] \[ B=\frac{\sqrt{3}}{5}\;G \]
Step 4: Final conclusion.
Hence, the magnitude of the earth's magnetic field is \[ \boxed{\frac{\sqrt{3}}{5}\;G} \]
The magnetic needle is free to rotate in a vertical plane parallel to the magnetic meridian.
Its north tip points downward at \(30^\circ\) with the horizontal.
Therefore, the angle of dip is \[ \delta=30^\circ \]
Step 2: Use the relation between total magnetic field and horizontal component.
If \(B\) is the magnitude of earth's magnetic field and \(H\) is its horizontal component, then \[ H=B\cos\delta \] Given, \[ H=0.3\;G \] and \[ \delta=30^\circ \]
Step 3: Substitute the values.
\[ 0.3=B\cos30^\circ \] \[ 0.3=B\left(\frac{\sqrt{3}}{2}\right) \] \[ B=\frac{0.3\times2}{\sqrt{3}} \] \[ B=\frac{0.6}{\sqrt{3}} \] \[ B=\frac{3}{5\sqrt{3}} \] \[ B=\frac{\sqrt{3}}{5}\;G \]
Step 4: Final conclusion.
Hence, the magnitude of the earth's magnetic field is \[ \boxed{\frac{\sqrt{3}}{5}\;G} \]
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