(a) Explanation Given that the speed of the proton is 4.0 × 10 7 m/s , the strength of Earth’s magnetic field is 1.0 × 10 − 6 T , the angle between the velocity of proton and the field direction is 25 ° . The perpendicular component of velocity of proton causes circular nature of motion and the parallel component of velocity causes the straight-line motion. In effect the proton moves in helical path in the field. The necessary centripetal force for the motion of the proton is provided by the magnetic force. m v ⊥ 2 r = e v ⊥ B (I) Here, m is the mass of proton, v ⊥ is the perpendicular component of velocity, r is the radius of the helix, e is the charge of proton, and B is the magnetic field. Rewrite the term v ⊥ in terms of the angle θ between the velocity and the field, and solve equation (I) for r . r = m v ⊥ e B = m v sin θ e B (II) Conclusion: Substitute 1.67 × 10 − 27 kg for m , 4.0 × 10 7 m/s for v , 25 ° for θ , 1 (b) To determine The pitch of the helix.

5th Edition

ISBN: 2818440038631

Author: GIAMBATTISTA

Publisher: MCG

See similar textbooks

Concept explainers

Magnetic Field Of Coaxial Cable

The word coaxial means same axis. So for a long straight cable to be coaxial, it must have concentric cylindrical shaped conductor around it. Coaxial cable (popularly called ‘coax’) has various applications ranging from current conductors to radiofrequen…

Videos

Question

Chapter 19, Problem 95P

(a)

To determine

The radius of the helix.

(b)

To determine

The pitch of the helix.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 19, Problem 94P

Chapter 19, Problem 96P

Students have asked these similar questions

= = You are preparing your house for a party with your classmates and friends, and want to set up an impressive light display to entertain them. From your study of fluids, you have come up with the idea based on the water flowing from the tank in the figure. You set up the tank as shown in the figure, filled to a depth h 1.15 m, and sitting on a stand of height { 0.300 m. You punch a hole in the tank at a height of Y1 = 0.102 m above the stand. (Ignore the thickness of the tank in your calculation.) You want to punch a second hole higher on the tank so that the streams of water from the two holes arrive at the same position on the table, in a catch basin at a distance d from the right edge of the stand. A pump will continuously carry water from the catch basin back up to the top of the tank to keep the water level fixed. Then, you will use laser pointers on the left side of the tank to light the two streams of water, which will capture the light (see the section on total internal…

A square metal sheet 2.5 cm on a side and of negligible thickness is attached to a balance and inserted into a container of fluid. The contact angle is found to be zero, as shown in Figure a, and the balance to which the metal sheet is attached reads 0.42 N. A thin veneer of oil is then spread over the sheet, and the contact angle becomes 180°, as shown in Figure b. The balance now reads 0.41 N. What is the surface tension of the fluid? N/m a

Sucrose is allowed to diffuse along a 12.0-cm length of tubing filled with water. The tube is 6.1 cm² in cross-sectional area. The diffusion coefficient is equal to 5.0 × 10-10 m²/s, and 8.0 × 10−14 x transported along the tube in 18 s. What is the difference in the concentration levels of sucrose at the two ends of the tube? .00567 kg is

Chapter 19 Solutions

PHYSICS

Ch. 19.2 - 19.2 An electron is moving with speed v in a...Ch. 19.2 - 19.1 Acceleration of Cosmic Ray Particle If v =...Ch. 19.2 - 19.2 Magnetic Force on an Electron Find the...Ch. 19.2 - Practice Problem 19.3 Velocity Component Parallel...Ch. 19.3 - 19.4 Ion Speed The magnetic field used in the mass...Ch. 19.3 - 19.5 Increasing Kinetic Energy in a Proton...Ch. 19.4 - 19.4 A particle’s helical motion is shown in Fig....Ch. 19.5 - 19.5 (a) , points east, and q is negative, so ...Ch. 19.5 - Practice Problem 19.6 Deflection of a Particle...Ch. 19.5 - Prob. 19.7PP

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.