Magnetic Force on Moving Charge 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 Moving Charge 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 | Magnetic Force on Moving Charge Numerical |
| Academic Session | 2025-2026 |
Magnetic Force on Moving Charge Numerical
Class-12 Nootan ISC Physics Solution Ch-7 Moving Charges and Magnetic Field.
Que-17: An electron, moving with a velocity of 5 x 10^7 m/s enters a magnetic field of 1 Wb m^-2 at an angle of 30°. Calculate the force on the electron. (ISC 2006)
Ans: F = qvB sinθ
=> 1.6 x 10^-19 x 5 x 10^7 x 1 x sin 30
=> 4 x 10^-12 N
Que-18: An electron is moving vertically upwards with a speed of 2.0 x 10^8 m s^-1. What will be the magnitude and direction of the force on the electron exerted by a horizontal magnetic field of 0.50 NA^-1 m^-1 directed towards west? What will be the acceleration of the electron?
Ans: F = qvB sin90
=> 1.6 x 10^-19 x 2.0 x 10^8 x 0.5 x 1
=> 1.6 x 10^-11 N due North
Que-19: A proton moving towards east in a horizontal plane enters a horizontal magnetic field of 0.34 T directed towards north with a speed of 2.0 x 10^-7 m/s. Calculate (i) the magnitude and direction of the force on the proton, (ii) radius of proton’s path and (iii) the lateral displacement of the proton while moving 0.20 m towards east.
Ans:
= 0.625 m
(iii) 0.20/2 x 10^7 = 10^-8 s
=> now acc in proton = qvB / m
=> 1.6 x 10^-19 x 2 x 10^7 x 0.34 x 10^-2 / 1.7 x 10^-27
=> 64 x 10^13 m/s
∴Lateral displacement s = 1/2 at²
=> 1/2 x 64 x 10^-13 x (10^-8)²
=> 32 x 10^-3 = 0.032 m
Que-20: A 2 MeV proton is moving perpendicular to a uniform magnetic field of 2.5 T. Find the force on the proton. Take mass of proton as 1.65 x 10^-27 kg.
Ans: KE of proton = 2 MeV
0.20/2 x 10^7 = 10^-8 s
=> now acc in proton = qvB / m
=> 1.6 x 10^-19 x 2 x 10^7 x 0.34 x 10^-2 / 1.7 x 10^-27
=> 64 x 10^13 m/s
∴Lateral displacement s = 1/2 at²
=> 1/2 x 64 x 10^-13 x (10^-8)²
=> 32 x 10^-3 = 0.032 m
Que-21: PQ is a long straight conductor carrying a current of 3A as shown in Fig. A proton moves with a velocity of 2 x 10^7 m/s parallel to it. Find the force acting on the proton. (ISC 2017)
Ans: Magnetic field due to wire of distance of 0.6 m
now F = qvB
=> 1.6 x 10^-19 x 2 x 10^7 x 10^-6
=> 3.2 x 10^-18 N
Que-22: An electron accelerated by a potential difference of 300 V, is moving at a distance of 4 mm from a long, straight wire. If a current of 5 A flows through the wire, what will be the force acting on the electron?
Ans: Energy acquired by electron = eV
— : End of Magnetic Force on Moving Charge Numerical Class-12 Nootan ISC Physics :–
Return to : – Nootan Solutions for ISC Class-12 Physics
Thanks
Please share with your friends
