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Ilya [14]
3 years ago
11

Particle 1 and particle 2 have masses of m1 = 2.2 × 10-8 kg and m2 = 4.8 × 10-8 kg, but they carry the same charge q. The two pa

rticles accelerate from rest through the same electric potential difference V and enter the same magnetic field, which has a magnitude B. The particles travel perpendicular to the magnetic field on circular paths. The radius of the circular path for particle 1 is r1 = 8 cm. What is the radius (in cm) of the circular path for particle 2?
Physics
1 answer:
Lorico [155]3 years ago
4 0

Answer:r_2=11.81 cm

Explanation:

Given

m_1=2.2\times 10^{-8} kg

m_2=4.8\times 10^{-8} kg

same charge on both masses

potential Energy due to Magnetic Field =Kinetic Energy of Particle

qV=\frac{mv^2}{2}

v=\sqrt{\frac{2qV}{m}}

and we know

Force due to magnetic field will Provide centripetal Force

qvB=\frac{mv^2}{r}

B=\frac{\sqrt{\frac{2Vm}{q}}}{r}

and B is equal for both particles

thus \frac{m}{r^2}=constant

\frac{m_1}{r_1^2}=\frac{m_2}{r_2^2}

r_2^2=\frac{4.8}{2.2}\times 8^2

r_2=11.81 cm

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