Q147 / 170 · Manipal (MET) 2020 · Physics
The horizontal component of earth's magnetic field at a place is $0.4 \times 10^4 \mathrm{~T}$. If the angle of dip is $45^{\circ}$ the value of total intensity of earth's magnetic field is
- Correct answerA$0.5 \times 10^{-4} \mathrm{~T}$
- B$0.4 \times 10^{-4} \mathrm{~T}$
- C$0.5 \times 10^{-6} T$
- D$0.4 \times 10^{-6} T$
Solution
Vertical component of earth's magnetic field is
$B_V=B_H \tan \delta$
where, $B_H=$ horizontal component of earth's magnetic field
$\delta=$ angle of dip.
Total intensity of earth's maanetic field
$\begin{aligned} B & =\sqrt{B_H^2+B_V^2} \\ & =\sqrt{B_H^2+B_H^2}\left[\because \tan \delta=\tan 45^{\circ}=1\right] \\ & =\sqrt{2} B_H \\ & =1.44 \times 0.4 \times 10^{-4} \mathrm{~T} \\ & =5.656 \times 10^{-5} \mathrm{~T} \\ & =0.5 \times 10^{-4} \mathrm{~T}\end{aligned}$
Hence, the correct answer is option: 'A'.