Magnetic Dipole Magnetic Moment 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.
Magnetic Dipole Magnetic Moment 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 Magnetic Dipole Magnetic Moment etc |
| Academic Session | 2025-2026 |
Numericals on Magnetic Dipole Magnetic Moment
Class-12 Nootan ISC Physics Solution Ch-9 Magnetic Field and Earth’s Magnetism.
Que-1. A small magnetic needle of magnetic moment 4.8 x 10^-2 J T^-1 is placed at 30° with a uniform external magnetic field B of magnitude 3.0 x 10^-2 T. What is the magnitude and direction of the torque acting on the needle?
Ans- τ = MB sinθ
=> 4.8 x 10^-2 x 3 x 10^-2 x 1/2
=> 7.2 x 10 ^-4 Nm
Magnet tends to become parallel to B
Que-2. A short bar-magnet of magnetic moment 0.9 JT^-1 placed with its axis at 45° with a uniform external magnetic field experiences a torque of magnitude 0.063 J. Find the strength of the magnetic field. What orientation of the bar-magnet corresponds to the stable equilibrium in the magnetic field?
Ans- τ = MB sinθ
B = τ/M sinθ
=> 0.063/(0.9 x 1/√2) = 0.99 T
Que-3. A bar-magnet placed in a uniform magnetic field of 0.3 T, with its axis at 30° to the field, experiences a torque of 0.06 N m. Find the magnetic moment of the bar-magnet.
Ans- τ = MB sinθ
M = τ/B sinθ
=> 0.06 x 2 /0.3 = 0.4 A m²
Que-4. A short bar-magnet of magnetic moment m = 0.32 J T^-1 is free to rotate in the plane of a uniform external magnetic field B of 0.15 T. What orientations of the magnet would correspond to its (i) stable and (ii) unstable equilibrium? What is the potential energy of the magnet in the magnetic field in each case?
Ans- for stable equilibrium θ = 0°
∴ U = -mB cos 0 = – mB = -0.32 x 0.15 = -0.048 J
and for unstable equilibrium θ = 180°
=> U = -mB cos 180° = mB = +0.048 J
Que-5. A bar-magnet of magnetic moment 1.5 A m², when suspended perpendicular to a magnetic field, is acted upon by a couple of 0.24 N-m (a) What is the magnitude of the magnetic field? (b) What will be the change in the potential energy of the magnet in coming from this position to the position of steady equilibrium? (c) If this magnet is equivalent to a current-carrying coil of 100 turns having a surface area of 150 cm², then what will be the current in the coil?
Ans-
(c) = m = niA
=> i = m/nA
=> 1.5 / 100 x 150 x 10^-4 = 1 A
Que-6. The magnetic moment of a short bar-magnet is 4.0 Am². Find the magnetic field at 5.0 cm from the centre of the magnet on its axis. If the magnetic field at a point situated on the equatorial line of this magnet be 0.05 T, then what is the distance of that point from the magnet?
Ans- Magnetic field due to short Bar magnet at its axis
Que-7. Two small magnets are placed along a straight line at a distance of 20 cm from each other and their north poles are facing each other. If their magnetic moments be 0.20 and 0.40 J T^-1, then find the magnetic field at the middle-point between them.
Ans-
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