Question

The horizontal component of Earth’s magnetic field at a place is 3 × 10^-5 T. If the dip at that place is 45°, the resultant magnetic field at that place is

• A. \(\frac{3}{\sqrt2}\times10^{-5}\ T\)
• B. \(\frac{3}{2}\sqrt3\times10^{-5}\ T\)
• C. \(3\sqrt2\times10^{-5}\ T\)
• D. \(3\times10^{-5}\ T\)

Answer 1

The Correct Option is C

Given: 
Horizontal component of Earth's magnetic field, \(B_H = 3 \times 10^{-5}\,T\)
Angle of dip, \( \delta = 45^\circ \)

Step-by-Step Explanation:

Step 1: Recall the relation between the resultant magnetic field (\( B \)), horizontal component (\( B_H \)), and angle of dip (\( \delta \)):

\[ B_H = B\cos\delta \]

Step 2: Solve for resultant magnetic field \( B \):

\[ B = \frac{B_H}{\cos\delta} \]

Step 3: Substitute given values:

\[ B = \frac{3 \times 10^{-5}}{\cos 45^\circ} \]

We know \(\cos 45^\circ = \frac{1}{\sqrt{2}}\). Thus:

\[ B = \frac{3 \times 10^{-5}}{\frac{1}{\sqrt{2}}} = 3\sqrt{2} \times 10^{-5}\,T \]

Final Conclusion:
The resultant magnetic field at the given place is \(3\sqrt{2}\times10^{-5}\,T\).