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3 years ago

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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 differenc

e 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?

2 answers:

marysya [2.9K]3 years ago

8 0

Answer:

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

blondinia [14]3 years ago

6 0

Answer:

Explanation:

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

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