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An electron of energy 1800 eV describes a circular path in magnetic field of flux density 0.4 T. The radius of path is (q = 1.6 x 10-19 C, me = 9.1 x 10-31 kg)
- a)2.58 x 10-4 m
- b)3.58 x 10-4 m
- c)2.58 x 10-3 m
- d)3.58 x 10-3 m
Correct answer is option 'B'. Can you explain this answer?
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An electron of energy 1800 eV describes a circular path in magnetic fi...
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An electron of energy 1800 eV describes a circular path in magnetic fi...
Given:
Energy of the electron, E = 1800 eV
Flux density of the magnetic field, B = 0.4 T
Charge of the electron, q = 1.6 x 10^-19 C
Mass of the electron, me = 9.1 x 10^-31 kg
To find:
Radius of the circular path, r
Formula used:
The equation for the radius of the circular path of a charged particle in a magnetic field is given by:
r = (me * v) / (q * B)
Where v is the velocity of the charged particle.
Derivation:
1. We can calculate the velocity of the electron using the formula:
E = (1/2) * me * v^2
Rearranging the equation, we get:
v^2 = (2 * E) / me
Substituting the given values, we have:
v^2 = (2 * 1800 eV) / (9.1 x 10^-31 kg) = 3.96 x 10^21 m^2/s^2
Taking the square root, we find:
v = 1.99 x 10^10 m/s
2. Now, substituting the values of me, v, q, and B into the formula for the radius of the circular path, we get:
r = (me * v) / (q * B)
= (9.1 x 10^-31 kg * 1.99 x 10^10 m/s) / (1.6 x 10^-19 C * 0.4 T)
= 3.58 x 10^-4 m
Therefore, the radius of the circular path of the electron is 3.58 x 10^-4 m, which corresponds to option B.
Energy of the electron, E = 1800 eV
Flux density of the magnetic field, B = 0.4 T
Charge of the electron, q = 1.6 x 10^-19 C
Mass of the electron, me = 9.1 x 10^-31 kg
To find:
Radius of the circular path, r
Formula used:
The equation for the radius of the circular path of a charged particle in a magnetic field is given by:
r = (me * v) / (q * B)
Where v is the velocity of the charged particle.
Derivation:
1. We can calculate the velocity of the electron using the formula:
E = (1/2) * me * v^2
Rearranging the equation, we get:
v^2 = (2 * E) / me
Substituting the given values, we have:
v^2 = (2 * 1800 eV) / (9.1 x 10^-31 kg) = 3.96 x 10^21 m^2/s^2
Taking the square root, we find:
v = 1.99 x 10^10 m/s
2. Now, substituting the values of me, v, q, and B into the formula for the radius of the circular path, we get:
r = (me * v) / (q * B)
= (9.1 x 10^-31 kg * 1.99 x 10^10 m/s) / (1.6 x 10^-19 C * 0.4 T)
= 3.58 x 10^-4 m
Therefore, the radius of the circular path of the electron is 3.58 x 10^-4 m, which corresponds to option B.
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An electron of energy 1800 eV describes a circular path in magnetic field of flux density 0.4 T. The radius of path is (q = 1.6 x 10-19 C, me= 9.1 x 10-31 kg)a)2.58 x 10-4mb)3.58 x 10-4 mc)2.58 x 10-3 md)3.58 x 10-3 mCorrect answer is option 'B'. Can you explain this answer? for NEET 2026 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about An electron of energy 1800 eV describes a circular path in magnetic field of flux density 0.4 T. The radius of path is (q = 1.6 x 10-19 C, me= 9.1 x 10-31 kg)a)2.58 x 10-4mb)3.58 x 10-4 mc)2.58 x 10-3 md)3.58 x 10-3 mCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for NEET 2026 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An electron of energy 1800 eV describes a circular path in magnetic field of flux density 0.4 T. The radius of path is (q = 1.6 x 10-19 C, me= 9.1 x 10-31 kg)a)2.58 x 10-4mb)3.58 x 10-4 mc)2.58 x 10-3 md)3.58 x 10-3 mCorrect answer is option 'B'. Can you explain this answer?.
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