The radius (r) of the circular orbit taken by a charged particle is related to its speed perpendicular to a magnetic field of strength B, and is given by

r = $\frac{mv}{qB}$ --------------(i)

Where,

q = charge of the particle

m = mass of the particle

Making v subject of the formula in equation (i) above gives

v = $\frac{qBr}{m}$ -------------------(ii)

Given;

r = 20cm = 0.2m

B = 0.3T

v = unknown

q = charge of proton = 1.6 x 10⁻¹⁹ C

m = mass of the proton = 1.67 x 10⁻²⁷kg

Substitute the values of m, q, B and r into equation (ii) above to get;

v = $\frac{1.6 * 10^{-19} * 0.3 * 0.2} {1.67*10^{-27} }$

Solving for v gives:

v = 5.75 x 10⁶ m/s

Therefore, the velocity of the proton is 5.75 x 10⁶ m/s

4 0

2 years ago

Which procedure would cause charged particles in a conductor to move in the same direction?

Alina [70]

The application of a potential difference across the conductor.

In fact, when the two ends of the conductor are at different voltage, the free positive charges in the conductor will start to flow toward the point at lower potential (while the free negative charges in the conductor will start to flow toward the point at higher potential), producing a net movement of charges, called "current".

7 0

3 years ago