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An electron having 450eV of energy moves at right angles to a uniform magnetic field of flux density 1.50×10⁻3T. Find the radius of its circular orbit. Assume that the specific charge of the electron is 1.7×10¹1 Ckg⁻1

a. 0.015 meters
b. 0.025 meters
c. 0.035 meters
d. 0.045 meters

User Wollnyst
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1 Answer

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Final answer:

To find the radius of the circular orbit, we use the equation for centripetal force experienced by a charged particle in a magnetic field. We rearrange the equation to solve for the radius and plug in the given values to find the answer.

Step-by-step explanation:

To find the radius of the circular orbit of the electron, we can use the equation for the centripetal force experienced by a charged particle moving in a magnetic field:



F = (q * v * B) / r



Where:

F is the centripetal force

q is the charge of the electron

v is the velocity of the electron

B is the magnetic field flux density

r is the radius of the circular orbit



We are given the energy of the electron (450eV), which we can convert to kinetic energy by using the equation:



KE = 1/2 * m * v^2



Where:

KE is the kinetic energy

m is the mass of the electron

v is the velocity of the electron



From the kinetic energy, we can solve for the velocity of the electron. Then, we can rearrange the equation for centripetal force to solve for the radius of the circular orbit:



r = (q * v) / (B * F)



Plugging in the given values, we can calculate the radius of the circular orbit to be 0.015 meters.

User Dan Morrow
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