Answered

Westonci.ca offers fast, accurate answers to your questions. Join our community and get the insights you need now. Explore our Q&A platform to find reliable answers from a wide range of experts in different fields. Experience the convenience of finding accurate answers to your questions from knowledgeable experts on our platform.

The magnetic field strength in an experiment is 5.0 x 10–4 Newtons per Amp meter (or 5.0 x 10–4 N A–1m–1), where the unit of force is the Newton (N = kg m s–2) and the unit of current is the Ampere (A = C s–1). A Coulomb, C, is a unit of charge. The electric field strength in another experiment is 1.8 x 104 N C–1. From these parameters, both the magnetic displacement and the electric displacement were each measured to be 6.0 cm. Calculate the velocity of the particles.

Sagot :

Cricetus

Answer:

The velocity of the particles is "0.36×10⁸ m/s".

Explanation:

The given values are:

Magnetic field strength,

B = 5.0×10⁻⁴ N/Amp

In another experiment,

F = 1.8×10⁴

By considering the Lorentz force, the velocity will be:

⇒  [tex]qE=qvB[/tex]

then,

⇒  [tex]v=\frac{F}{B}[/tex]

On substituting the given values, we get

⇒     [tex]=\frac{1.8\times 10^4}{5.0\times 10^{-4}}[/tex]

⇒     [tex]=0.36\times 10^8 \ m/s[/tex]

Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. Stay curious and keep coming back to Westonci.ca for answers to all your burning questions.