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 \(3\sqrt{2}\times10^{-5}\,T\).

Answer 2

Let the horizontal component of Earth's magnetic field be $B_H$ and the vertical component be $B_V$. From the given data, we know that $B_H = 3 \times 10^{-5}$ T. The dip $\delta = 45^\circ$, so we can use the formula for the resultant magnetic field $B$ as: \[ B = \sqrt{B_H^2 + B_V^2} \] Also, the vertical component $B_V$ is related to the horizontal component by the angle $\delta$: \[ B_V = B_H \tan \delta \] Substituting $\delta = 45^\circ$ and $\tan 45^\circ = 1$, we get: \[ B_V = B_H \] Now, we substitute into the formula for the resultant magnetic field: \[ B = \sqrt{B_H^2 + B_H^2} = \sqrt{2 B_H^2} = B_H \sqrt{2} \] Substitute the value of $B_H$: \[ B = 3 \times 10^{-5} \times \sqrt{2} = 3 \sqrt{2} \times 10^{-5} \, \text{T} \] Thus, the resultant magnetic field is $3 \sqrt{2} \times 10^{-5}$ T.