Home
Class 12
PHYSICS
A short bar magnet placed in a horizonta...

A short bar magnet placed in a horizontal plane has its axis aligned along the magnetic north south direction. Null points are found on the axis of the magnet at `14cm` from the centre of the magnet. The earth's magnetic field at the plane is `0*36G` and the angle of dip is zero. What is the total magnetic field on the normal bisector of the magnet at the same distance as the null points (i.e. 14 cm) from the centre of the magnet? (At null points, field due to a magnet is equal and apposite to the horizontal component of earth's magnetic field).

Text Solution

AI Generated Solution

To solve the problem, we need to determine the total magnetic field on the normal bisector of the magnet at a distance of 14 cm from the center of the magnet. Here’s a step-by-step solution: ### Step 1: Understand the Problem We have a short bar magnet aligned along the magnetic north-south direction, and we know the null points are located at 14 cm from the center of the magnet. The Earth's magnetic field is given as 0.36 G, and the angle of dip is zero, indicating that the magnetic field is horizontal. **Hint:** Identify the key parameters: distance from the center of the magnet (14 cm), Earth's magnetic field (0.36 G), and the nature of the magnetic field at null points. ### Step 2: Determine the Magnetic Field at Null Points ...
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • MAGNETIC EFFECT OF CURRENT AND MAGNETISM

    PRADEEP|Exercise Conceptual Problems (d)|2 Videos

  • MAGNETIC EFFECT OF CURRENT AND MAGNETISM

    PRADEEP|Exercise Very Short Answer Questions (a)|2 Videos

  • ELECTROSTATICS

    PRADEEP|Exercise ASSERTION-REASON TYPE QUESTIONS|2 Videos

  • OPTICS

    PRADEEP|Exercise Multiple choice questions|1 Videos

Similar Questions

Explore conceptually related problems

A short bar magnet placed in a horizontal plane has its axis aligned along the magnetic north south direction. Null points are found on the axis of the magnet at 14 cm from the centre of the magnet. The earth's magnetic field at the place is 0.36 G and the angle of dip is zero. What is the total magnetic field on the normal bisector of the magnet at the same distance as the null point (i.e., 14 cm) from the centre of the magnet ? (At null points, field due to a magnet is equal and opposite to the horizontal component of earth's magnetic field.)

A short bar magnet is placed in a horizontal plane with its axis in the magnetic merdian . Null points are found on its equitorial line (i.e., its normal bisector) at 12.5 G and the angle of dip is zero. (i) What is the total magnetic field at points on the axis of the magnet located by the same distance (12.5 cm) as the null-points from the centre? (ii) Locate the null points when the magnet is turned around by 180^(@) . Assume that the length of the magnet is negligible as compared to the distance of the null-point from the centre of the magnet.

Knowledge Check

  • A very small magnet is placed in the magnetic meridian with its south pole pointing north. The null point is obtained 20 cm away from the centre of the magnet. If the earth's magnetic field (horizontal component) at this point be 0.3 Gauss, the magnetic moment of the magnet is

    A
    `8.0xx10^(2) e.m.u.`
    B
    `1.2xx10^(3) e.m.u.`
    C
    `2.4xx10^(3) e.m.u.`
    D
    `3.6xx10^(3) e.m.u.`

  • At certain place, the horizontal component of earth's magnetic field is 3.0G and the angle dip at the place is 30^(@) . The magnetic field of earth at that location

    A
    4.5G
    B
    5.1G
    C
    3.5G
    D
    6.0G

  • A short bar magnet of magnetic moment 1.2Am^(2) is placed in the magnetic meridian with its south pole pointing the north. If a neutral point is found at a distance of 20cm from the centre of the magnet, the value of the horizontal component of the earth's magnetic field is

    A
    `3xx10^(-5)T`
    B
    `3xx10^(-4)T`
    C
    `3xx10^(3)T`
    D
    `3xx10^(-2)T`

    • Similar Questions

    Explore conceptually related problems

    In the magnetic meridian of a certain place, the horizontal component of the earth's magnetic field is 0*26G and dip angle is 60^@ . What is the magnetic field of earth at this location?

    Calculate earth's magnetic field at a place where angle of dip delta is 60^(@) and horizontal component of earth's magnetic field is 0.3 G

    A 30 cm long bar magnet is placed in the magnetic meridian with its north pole pointing south. The neutral point is obtained at a distance of 40 cm from the centre of the magnet. Pole strength of the magnet is (The horizontal component of earth’s magnetic field is 0.34 Gauss)

    A short bar magnet is placed horizontally in the magnetic meridian with its north pole pointing north and the neutral points are found 0.15m from the centre of the magnet. Find the distance between the neutral points if the magnet is reversed.

    At a place, the horizontal component of earth's magnetic field is B and angle of dip is 60^@ . What is the value of horizontal component of the earth's magnetic field at equator?

    Home
    Profile