Magnetic Force on Current Carrying Conductor Numerical Class-12 Nootan ISC Physics Solution Ch-7 Moving Charges and Magnetic Field. 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 Force on Current Carrying Conductor Numerical Class-12 Nootan ISC Physics Solution Ch-7 Moving Charges and Magnetic Field
| Board | ISC |
| Class | 12 |
| Subject | Physics |
| Book | Nootan |
| Chapter-7 | Moving Charges and Magnetic Field |
| Topics | Numericals on Magnetic Force on Current Carrying Conductor |
| Academic Session | 2025-2026 |
Numericals on Magnetic Force on Current Carrying Conductor
Class-12 Nootan ISC Physics Solution Ch-7 Moving Charges and Magnetic Field.
Que-29: Find the force per unit length on a wire carrying a current of 8 A and making 30° angle with the direction of a uniform magnetic field of 0.15 T.
Ans: F = iBl sinθ
=> 8 x 0.15 x 1 x sin 30
=> 0.6 N/m
Que-30: A current of 5.0 A is flowing upwards in a long vertical wire placed in a uniform horizontal northward magnetic field of 0.020 T. How much force and in what direction will the field exert on 0.06 m length of the wire?
Ans: F = iBl sinθ
=> 5 x 0.02 x 0.06 x sin 90
=> 6 x 10^-3 N/m
directions can be obtained by Fleming left hand rule
Que-31: Two parallel rails, distant 2.0 × 10^-2 m, are laid in north-south direction. On these rails is kept a metal cylinder of mass 4.0 x 10^-2 kg. A battery is connected to the rails with its positive terminal joined to the rail on the eastern side. The battery sends a current of 3.0 A in the cylinder. If a uniform magnetic field of 1.2 T directed upwards be there, then find out the magnitude and direction of the magnetic force imposed upon the cylinder. What would be the acceleration in the cylinder?
Ans- F = iBl sinθ
=> 3 x 1.2 x 2.0 x 10^-2 sin 90
=> 7.2 x 10^-2 N
acc = F/m = 7.2 x 10^-2 / 4.0 x 10^-2
=> 1.8 m/s²
directions can be obtained by Fleming left hand rule
Que-32: On a 30° inclined smooth plane, a thin current-carrying rod is placed parallel to the horizontal ground. The plane is located in a uniform vertical magnetic field of 0.15 T. The mass/length of the rod is 0.30 kg/m. Find the current required to keep the rod stationary.
Ans:
Que-33: A current-carrying conductor has 8.0 x 10^22 free electrons per metre length and their mean drift velocity is 8.0 x 10^-5 m/s . If a magnetic field of 0.10 T be applied perpendicular to the conductor, then calculate (i) current in the conductor, (ii) force on an electron and (iii) force imposed per metre length of the conductor.
Ans:
Que-34: A 60 cm long wire (mass 10 g) is hanged by two flexible wires in a magnetic field of 0.40 T. Find the magnitude and direction of the current required to be flown to neutralize the tension of the hanging wires.
Ans: for neutralising tension
mg = iBl
=> 10 x 10^-2 x 9.8 = i x 0.4 x 0.6
=> i = 10 x 9.8 x 10^-2 / 0.4 x 0.6
=> 0.41 A
Force should be upward therefore direction of current will be left to right
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