Path of Charged Particle in Uniform Magnetic Field Numerical Class-12 Nootan ISC Physics 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.
Path of Charged Particle in Uniform Magnetic Field Numerical Class-12 Nootan ISC Physics 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 Path of Charged Particle in Uniform Magnetic Field |
| Academic Session | 2025-2026 |
Numericals on Path of Charged Particle in Uniform Magnetic Field
Class-12 Nootan ISC Physics Ch-7 Moving Charges and Magnetic Field
Que-23: A uniform magnetic field of 6.5 G (1 G = 10^-4 T) is maintained in a chamber. An electron is shot into the field with a speed of 4.8 × 10^6 m/s normal to the field. Determine the radius of the circular orbit and the frequency of revolution of the electron in its orbit. Does the frequency depend upon the speed of the electron?
Ans: r = m v / q B
=> 9.1 x 10^-31 x 4.8 x 10^6 / 1.6 x 10^-19 x 6.5 x 10^-4
=> 4.2 x 10^-2 m = 4.2 cm
T = 2πm / qB
=> 2 x 3.14 x 9.1 x 10^-31 / 1.6 x 10^-19 x 6.5 x 10^-4
f = 1/T
=> 1.6 x 10^-19 x 6.5 x 10^-4 / 2 x 3.14 x 9.1 x 10^-31
=> 18 x 10^6 Hz
frequency is independent of velocity
Que-24: An α-particle is moving along a circle of radius 0.45 m in a magnetic field of 12 T. Find (i) the speed of the particle, (ii) the time-period of the particle. (Mass of a-particle is 6.7 x 10^-27 kg.)
Ans: r = m v / q B
=> v = q B r / m
=> 3.2 x 10^-19 x 12 x 0.45 / 6.7 x 10^-27
=> 2.58 x 10^8 m/s
T = 2πm / qB
=> 2 x 3.14 x 6.7 x 10^-27 / 3.2 x 10^-19 x 12
=> 1.09 x 10^-8 s
Que-25: An electron is moving with a speed of 3.0 x 10^7 ms^-1 along a circle in a magnetic field of 0.50 N A^-1 m^-1. Calculate the radius of the circle and the kinetic energy of the electron.
Ans: r = m v / q B
=> 9.1 x 10^-31 x 3 x 10^7 / 1.6 x 10^-19 x 0.5
=> 34 x 10^-5 m = 0.34 mm
K = 1/2 mv²
=> 1/2 x 9.1 x 10^-31 x (3 x 10^7)²
=> 9.1 x 9/2 x 10^-17
=> 4.1 x 10^-16 J
Que-26: An electron of 40 eV energy is revolving in a circular path in a magnetic field of 9 x 10^-5 T. Determine: (i) speed of the electron, (ii) radius of the circular path.
Ans:
Que-27: The velocity of an electron at a point A is 10^7 m s^-1. Find: (i) direction and magnitude of the magnetic field due to which the electron will go from A to B the electron in going from A to B on the semi-circular path shown, (ii) time taken by the electron going from A to B.
Ans:
Since electron is moving clock wise therefore magnetic field is vertically down ward
Que-28: A proton, a deuteron and an α-particle, whose kinetic energies are same, enter perpendicularly a magnetic field. Compare the radii of their circular paths.
Ans:
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