At a certain location in Africa, compass points $12^{0}$ west of geographic north. The north tip of magnetic needle of a dip circle placed in the plane of magnetic meridian points $60^{0}$ above the horizontal. The horizontal component of earth's field is measured to be $0.16$ G. The magnitude of earth's field at the location is :

0.32

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

0.42

No worries! We‘ve got your back. Try BYJU‘S free classes today!

4.2

No worries! We‘ve got your back. Try BYJU‘S free classes today!

3.2

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $0.32$ G
Here, $H_{E} = 0.16 G = 0.16 \times 10^{- 4} T$ , dip angle $(δ) = 60^{0}$
Then, magnitude of earth's field.
$B_{E} = \frac{H_{E}}{c o s δ} = \frac{0.16 \times 10^{- 4}}{c o s 60^{0}} T \Rightarrow B_{E} = \frac{0.16 \times 10^{- 4}}{1 / 2} = 0.32 \times 10^{- 4} T = 0.32 G$

Suggest Corrections

Similar questions

Q. At a certain location in Africa, a compass points

12^{\circ}

west of the geographic north. The north tip of the magnetic needle of a dip circle placed in the plane of magnetic meridian points

60^{\circ}

above the horizontal. The horizontal component of the earth's field is measured to be

0.16 G .

What is the direction and magnitude of the Earth's field at the location?

A magnetic needle free to rotate in a vertical plane parallel to the magnetic meridian has its north tip pointing down at 22º with the horizontal. The horizontal component of the earth’s magnetic field at the place is known to be 0.35 G. Determine the magnitude of the earth’s magnetic field at the place.

Join BYJU'S Learning Program

The correct option is A 0.32GHere, HE=0.16G=0.16×10−4T, dip angle (δ)=600Then, magnitude of earth's field.BE=HEcosδ=0.16×10−4cos600T⇒BE=0.16×10−41/2=0.32×10−4T=0.32G