A particle with a mass of 6.64 * 10-27 kg and a charge of + 3.20 * 10-19 C is accelerated from rest through a potential difference of 2.45 * 106 V. The particle then enters a uniform 1.60-T magnetic field. If the particle's velocity is perpendicular to the magnetic field at all times, what is the magnitude of the magnetic force exerted on the particle?

Given that,

Mass m = 6.64*10^-27kg

Charge q = 3.2*10^-19C

Potential difference V = 2.45*10^6V

Magnetic field B = 1.6T

The force in a magnetic field is given as Force = q• (V*B)

Since V and B are perpendicular i. e 90°

Force = q•V•BSin90

F=q•V•B

So we need to find the velocity

Then, K•E is equal to work done by charge I. e K•E=U

K•E = ½mV²

K•E = ½ * 6.64*10^-27 V²

K•E = 3.32*10^-27 V²

U = q•V

U = 3.2*10^-19 * 2.45*10^6

U = 7.84*10^-13

Then, K•E = U

3.32*10^-27V² = 7.84*10^-13

V² = 7.84*10^-13 / 3.32*10^-27

V² = 2.36*10^14

V=√2.36*10^14

V = 1.537*10^7 m/s

So, applying this to force in magnetic field

F=q•V•B

F = 3.2*10^-19 * 1.537*10^7 * 1.6

F = 7.87*10^-12 N

F = 7.86 * 10^-12 N

Explanation:

Given:-

- The mass of the particle, m = 6.64 * 10^-27 kg

- The charge of the particle, q = + 3.20 * 10^-19 C

- The potential difference applied, ΔV = 2.45 * 10^6 V

- The strength of magnetic field, B = 1.60 T

Find:-

What is the magnitude of the magnetic force (F) exerted on the particle?

Solution:-

- To determine speed (v) of the accelerated particle under potential difference (ΔV) we will use the energy-work principle. Where work is done (U) by the applied potential difference to accelerate the particle (Δ K. E)

Δ K. E = U

0.5*m*v^2 = ΔV*q

v = √ (2ΔV*q / m)

- The Lorentz force (F) exerted by the magnetic field (B) on the charge particle (q) with velocity (v) is given by:

F = q*v*B

F = q*√ (2ΔV*q/m) * B

F = (+3.20 * 10^-19) * √ (2 (2.45 * 10^6) * (+3.20 * 10^-19) / 6.64 * 10^-27) * (1.60)

F = 7.86 * 10^-12 N