A Particle whose is twice the mass of Proton and Charge is four times the charge of proton is allowed through a uniform magnetic field field of strength 1.7 mT and is pespendicular to the velocity of the particle find the radius and time period of revolution of the particle if the velocity of the particle is 2.5X10^4m/s.
Answers
Answer:
r=0.5678m,w=0.8×10^-6
Explanation:
r=mv/qB | w=2πm/qB
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put value and solve the question
The radius of the particle is 7.4 mm.
The time period of the revolution of the particle is 1.858 × 10^-6 s.
Given: The mass of a particle is twice the mass of a Proton and the charge is four times the charge of the proton. The uniform magnetic field is 1.7mT. The velocity of the particle is 2.5 X 10^4 m/s.
To Find: The radius and time period of revolution of the particle.
Solution:
We know that the radius can be found using the formula,
Radius r = mv / qB ...(1)
where m = mass of the particle, q = charge of the particle, B = magnetic field strength, v = velocity of the particle.
- We also know that the formula for finding the time period of revolution can be given by the formula,
Time period T = 2πr / v ...(2)
where r = radius of the particle, v = velocity of the particle.
Coming to the numerical,
Mass of proton = 1.6 × 10^-27 kg
So, mass of the particle (m) = 2 × 1.6 × 10^-27 kg = 3.2 × 10^-27 kg
Charge of proton = 1.6 × 10^-19 C
So, charge of the particle (q) = 4 × 1.6 × 10^-19 C = 6.4 × 10^-19 C
Magnetic field strength (B) = 1.7 mT
The velocity of the particle (v) = 2.5 × 10^4 m/s.
Putting respective values in (1), we get;
Radius r = mv / qB
⇒ r = ( 3.2 × 10^-27 × 2.5 × 10^4 ) / ( 6.4 × 10^-19 × 1.7 × 10^-3 )
⇒ r = 8 × 10^-23 / 1.088 × 10^-21
⇒ r = 0.0074 m
= 7.4 mm
Now, to find the time period of revolution of the particle we use (2). Putting respective values in (2) we get,
Time period T = 2πr / v
⇒ T = ( 2 × 3.14 × 0.0074 ) / 2.5 × 10^4
⇒ T = 1.858 × 10^-6 s
Hence, the radius of the particle is 7.4 mm
the time period ofthe revolution of the particle is 1.858 × 10^-6 s.
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