A magnetic needle free to rotate in a vertical plane, orients itself with its axis vertical at a certain place on the Earth. What is the value of Angle of dip at this place?

80^{o}

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

90^{o}

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

190^{o}

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

0^{o}

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

Open in App

Solution

The correct option is B $90^{o}$
Since the magnetic needle orients itself with its axis vertical, thus all the magnetic field must be in vertical direction and so horizontal component of earth's magnetic field must be zero.
Thus, angle of dip at that place is $90^{o}$

Suggest Corrections

Similar questions

Q. A magnetic needle free to rotate in a vertical plane, orients itself with its axis vertical at a certain place on the Earth. What is the value of Horizontal component of Earth's field?

Q. A magnetic needle is free to rotate in vertical plane. Initially plane of rotation of needle is perpendicular to magnetic meridian and needle is in vertical position (i.e. angle of dip is

90^{\circ}

). Now if plane of rotation of needle is rotated by

60^{\circ}

about a vertical axis and angle of dip measured in this position is found to be

30^{\circ}

, then true angle of dip at this place is

Q. (a) Small compass needle of magnetic moment 'm' is free to turn about an axis perpendicular to the direction of uniform magnetic field 'B'. The moment of inertia of the needle about the axis is 'I'. The needle is slightly disturbed from its stable position and then released. Prove that it executes simple harmonic motion. Hence deduce the expression for its time period.
(b) A compass needle , free to turn in a vertical plane orients itself with its axis vertical at a certain place on the earth. find out the values of (i) horizontal component of earth's magnetic field and (ii) angle of dip at the place.

Q. A magnetic needle freed to rotate around a fixed vertical plane stay at an angle

60^{0}

with the horizontal . If the dip at that place is

37^{0}

, find the angle of the fixed vertical plane with the meridian.

A magnetic needle is free to rotate in a vertical plane which makes an angle of $60^{0}$ with the magnetic meridian. If the needle stays in a direction making an angle of tan $^{- 1} (\frac{2}{\sqrt{3}})$ with the horizontal, what would be the dip at that place ?

Join BYJU'S Learning Program

A magnetic needle free to rotate in a vertical plane, orients itself with its axis vertical at a certain place on the Earth. What is the value of Angle of dip at this place?

The correct option is B 90oSince the magnetic needle orients itself with its axis vertical, thus all the magnetic field must be in vertical direction and so horizontal component of earth's magnetic field must be zero.Thus, angle of dip at that place is 90o

The correct option is B $90^{o}$
Since the magnetic needle orients itself with its axis vertical, thus all the magnetic field must be in vertical direction and so horizontal component of earth's magnetic field must be zero.
Thus, angle of dip at that place is $90^{o}$