Torque on Current Carrying Coil in Magnetic Field Numerical Class-12 Nootan ISC Physics Solution Ch-8 Torque on a Current-Loop : Moving coil Galvanometer. 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.
Torque on Current Carrying Coil in Magnetic Field Numerical Class-12 Nootan ISC Physics Solution Ch-8 Torque on a Current-Loop : Moving coil Galvanometer
| Board | ISC |
| Class | 12 |
| Subject | Physics |
| Book | Nootan |
| Chapter-8 | Torque on a Current-Loop : Moving coil Galvanometer. |
| Topics | Torque on Current Carrying Coil in Magnetic Field Numerical |
| Academic Session | 2025-2026 |
Torque on Current Carrying Coil in Magnetic Field Numerical
Que-1. When a coil of magnetic moment 2.5 × 10^-8 J T^-1 is placed with its face parallel to a uniform magnetic field, it experiences a torque of 7.5 x 10^9 J. Find the magnitude of the field.
Ans- When face of coil is parallel to MF its area vector makes angle 90° with MF and coil experiences max torque
∴(τ)max = MB sin 90°
=> B = (τ)max/M
=> 7.5 x 10^-9 / 2.5 x 10^-8 = 0.3 T
Que-2. A square coil of side 10 cm consists of 20 turns and carries a current of 12 A. It is suspended vertically in a uniform horizontal magnetic field of 0.80 T, such that the normal to the plane of the coil makes an angle of 30° with the field direction. Find the magnitude of the torque experienced by the coil.
Ans- τ = MB sin θ {M = NiA}
=> NiAB sinθ
=> 20 x 12 x (10 x 10^-2)² x 0.8 x sin 30
=> 20 x 12 x 10^-2 x 0.8 x 1/2 = 0.96 Nm
Que-3. A rectangular loop of area 5 m², has 50 turns and carries a current of 1 A. It is held in a uniform magnetic field of 0.1 T, at an angle of 30°. Calculate the torque experienced by the coil. (ISC 2019)
Ans- τ = NiAB sinθ
=> 50 x 1 x 5 x 0.1 x sin 60 {angle from area vector}
=> 25√3/2 = 21.65 Nm
Que-4. A circular coil of radius 6.0 cm and 25 turns carries a current of 10 A. It is suspended vertically in a uniform magnetic field of 1.2 T and the field lines are horizontal in the plane of the coil. Compute torque acting on the coil.
Ans- τ = NiAB sinθ
=> 25 x 10 x (π x 6 x 10^-2)² x 1.2 x sin 90
=> 3.4 Nm
Que-5. A rectangular coil of sides 8.0 cm and 6.0 cm having 2000 turns and carrying a current of 0.20 A is suspended in a uniform magnetic field of 0.20 T directed along the positive X-axis. (a) What is the net force on the coil? (b) What is the maximum torque the coil can experience and in which orientation? (c) For which orientation of the coil the torque is zero? When is this equilibrium stable and when is it unstable?
Ans- (a) Net force on a closed loop in an uniform magnetic field is Zero
(b) τ = NiAB sinθ
=> 2000 x 0.2 x (0.08 x 0.06) x 0.20
=> 0.384 N m , when the X-axis lies in the plane of the coil,
(c) Torque is zero when the coil lies in the Y-Z-plane. Equilibrium is stable when the area vector A is parallel to B and unstable when A is anti-parallel to B.
Que-6. A coil of area 5 cm² and having 100 turns is placed in a uniform magnetic field of 1.5 T. When a current of 0.2 A is passed through the coil find the magnetic dipole moment of the coil and the maximum torque produced.
Ans- Magnetic Dipole Moment
m = NiA
=> 100 x 0.2 x 5 x 10^-4 = 1 x 10^-2 A m²
(τ)max = mB sin90
=> 1 x 10^-2 x 1.5 = 1.5 x 10^-2 Nm
Que-7. A uniform magnetic field of 3000 gauss is established along the positive Z-axis. A rectangular loop of size 10 cm x 5 cm carries a current of 12 A. What is the magnitude and direction of the torque on the loop in the four cases shown in the figure?
Ans-
— : End Torque on Current Carrying Coil in Magnetic Field Numerical Class-12 Nootan ISC Physics Solution Ch-8 Torque on a Current-Loop : Moving coil Galvanometer :–
Return to : – Nootan Solutions for ISC Class-12 Physics
Thanks
Please share with your friends
