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| Numerical 13 |
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| 13. An electron is accelerated from rest through a potential difference of 3 KV. It enters into the region of a uniform, perpendicular magnetic field of 0.2T. What is the radius of the path of electron inside the field? |
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| Suggested solution: |
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| Let v be the speed acquired by the electron when accelerated by a potential difference of V. |
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| Then, |
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| (1/2) mv2 = eV |
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| or v = 2eV/m |
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| The radius R of the circular path inside the magnetic field is |
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| mv2 / R = BeV |
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| Given: B = 0.2 T, V = 3000 V, e = 1.6 x 1019 c, m = 9.1 x 1031 kg |
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| = R = (1/0.2) 2 x 9.1 x 10-31 x 3 x 103 /1.6 x 1019 (m) |
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| = 2.925 x 10-2m = 2.295cm |
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Moving Charges and Magnetism
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