This page focuses on the detailed Moving Charges and Magnetism question answers for Class 12 Physics Moving Charges and Magnetism, addressing the question: 'A magnetic field set up using Helmholtz coils (described in Exercise 4.16) is uniform in a small region and has a magnitude of 0.75 T. In the same region, a uniform electrostatic field is maintained in a direction normal to the common axis of the coils. A narrow beam of (single species) charged particles all accelerated through 15 kV enters this region in a direction perpendicular to both the axis of the coils and the electrostatic field. If the beam remains undeflected when the electrostatic field is 9.0 x 10-5 V m-1, make a simple guess as to what the beam contains. Why is the answer not unique?'. The solution provides a thorough breakdown of the question, highlighting key concepts and approaches to arrive at the correct answer. This easy-to-understand explanation will help students develop better problem-solving skills, reinforcing their understanding of the chapter and aiding in exam preparation.

Question 20

A magnetic field set up using Helmholtz coils (described in Exercise 4.16) is uniform in a small region and has a magnitude of 0.75 T. In the same region, a uniform electrostatic field is maintained in a direction normal to the common axis of the coils. A narrow beam of (single species) charged particles all accelerated through 15 kV enters this region in a direction perpendicular to both the axis of the coils and the electrostatic field. If the beam remains undeflected when the electrostatic field is 9.0 x 10^{-5} V m^{-1}, make a simple guess as to what the beam contains. Why is the answer not unique?

Answer

Magnetic field, B = 0.75 T

Accelerating voltage, V = 15 kV = 15 × 103 V

Electrostatic field, E = 9 × 105 V m - 1

Mass of the electron = m

Charge of the electron = e

Velocity of the electron = v

Kinetic energy of the electron = eV

Since the particle remains undeflected by electric and magnetic fields, we can infer that the electric field is balancing the magnetic field.

Putting equation (2) in equation (1), we get

This value of specific charge e/m is equal to the value of deuteron or deuterium ions. This is not a unique answer. Other possible answers are He^{++}, Li^{++}, etc.

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Nidhi
2018-01-18 18:51:52

Answer is correct but mentor forgot to mention proper signs while writing statements in the questions.

Simar
2018-01-18 17:31:14

Your solution has wrong calculation. See the value of electric field in the question and what you have put in the answer. See the power of 10 its -5 not 5

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