Earth Magnetism Numerical Class-12 Nootan ISC Physics Solution Ch-9 Magnetic Field and Earth’s Magnetism. Step by step solutions of Kumar and Mittal Physics of Nageen Prakashan as council latest prescribe guideline for upcoming exam. Visit official Website CISCE for detail information about ISC Board Class-12 Physics.
Earth Magnetism Numerical Class-12 Nootan ISC Physics Solution Ch-9 Magnetic Field and Earth’s Magnetism
| Board | ISC |
| Class | 12 |
| Subject | Physics |
| Book | Nootan |
| Chapter-9 | Magnetic Field and Earth’s Magnetism |
| Topics | Numericals on Earth Magnetism |
| Academic Session | 2025-2026 |
Numericals on Earth Magnetism
Class-12 Nootan ISC Physics Solution Ch-9 Magnetic Field and Earth’s Magnetism
Que-8. Two mutually perpendicular lines are drawn on a table. Two small magnets, whose magnetic moments are 0.108 and 0.192 JT^-1, are placed on these lines. If the distance of the point of intersection of these lines be respectively 30 and 40 cm from these magnets, then find the resultant magnetic field at the point of intersection.
Ans-
Que-9. The horizontal component of earth’s magnetic field at a place is √3 times the vertical component. What is angle of dip at that place?
Ans- tan θ = BV/BH θ = Angle of Dip
=> tan θ = BV/√3 BV = 1/√3
=> θ = 30°
Que-10. If the ratio of the horizontal component of earth’s magnetic field to the resultant magnetic field at a place is 1/√2, what is the angle of dip at that place?
Ans- BV = Be sin θ
=> sin θ = BV/Be = 1/√2
=> θ = 45°
Que-11. A magnetic needle free to rotate in a vertical plane parallel to the magnetic meridian has its north tip pointing down at 60° with the horizontal. The horizontal component of earth’s magnetic field is 0.3 x 10^-4 Wb/m². Find the vertical component and total magnetic field of earth.
Ans-
Que-12. A magnetic needle free to rotate about the vertical direction (compass) points 3.5° west of the geographic north. Another magnetic needle free to rotate in a vertical plane parallel to the magnetic meridian (dip needle) has its north tip pointing down at 18° with the horizontal. The horizontal component of earth’s magnetic field at the place is 0.40 G. Find the magnitude and direction of the earth’s magnetic field BE at the place. (tan 18° = 0.325).
Ans-
Que-13. A short bar-magnet is placed with its north pole pointing north. The neutral point is 10 cm away from the centre of the magnet. If BH = 0.4 G, calculate the magnetic moment of the magnet.
Ans- When north of magnet faces north Neutral point occur at equitorial line
i.e. BH = μο m/4π r³
=> 0. 4x 10^-4 = 10^-7 x m/(0.1)³
=> m = 0.4 A m²
Que-14. The magnetic moment of a small magnet is 4.8 A m². It is kept in earth’s magnetic field in magnetic meridian with its north pole pointing towards geographical south. The neutral points are situated at a distance of 30 cm from the centre of the magnet. Calculate the horizontal component of the earth’s magnetic field.
Ans- For north pole facing south pole of earth Neutral point will occur at axial line
∴ BH = μο 2m/4π r³
=> 10^-7 x 2 x 4.8 / (0.3)³
=> 0.35 x 10^-4 T
=> 0.35 G
Que-15. A small bar-magnet is placed in a horizontal plane with its axis in the magnetic meridian. Neutral points are obtained on its equatorial line at 12.5 cm from the centre of the magnet. The earth’s magnetic field at the place is 0.38 G and the angle of dip is zero. (a) Find the total magnetic field on the axis of the magnet at 12.5 cm from the centre of the magnet. (b) Locate the neutral points when the magnet is turned around by 180°.
Ans- Let the magnetic dipole moment of magnet is m
∴ Net Magnetic field at axial line at distance 12.5 cm
— : End of Magnetic Dipole Magnetic Moment Numerical Class-12 Nootan ISC Physics Solution Ch-9 Magnetic Field and Earth’s Magnetism. :–
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