Question

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.

Answer 1

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:

⇒  $qE=qvB$

then,

⇒  $v=\frac{F}{B}$

On substituting the given values, we get

⇒     $=\frac{1.8\times 10^4}{5.0\times 10^{-4}}$

⇒     $=0.36\times 10^8 \ m/s$