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Question: A proton (mass \(= 1.67 \times 10 ^ { - 27 } \mathrm {~kg}\) and charge \(3.4 \times 10 ^ { 7 } \mat...

A proton (mass =1.67×1027 kg= 1.67 \times 10 ^ { - 27 } \mathrm {~kg} and charge 3.4×107 m/sec3.4 \times 10 ^ { 7 } \mathrm {~m} / \mathrm { sec }. The acceleration of the proton should be

A

6.5×1015 m/sec26.5 \times 10 ^ { 15 } \mathrm {~m} / \mathrm { sec } ^ { 2 }

B

6.5×1013 m/sec26.5 \times 10 ^ { 13 } \mathrm {~m} / \mathrm { sec } ^ { 2 }

C

6.5×1011 m/sec26.5 \times 10 ^ { 11 } \mathrm {~m} / \mathrm { sec } ^ { 2 }

D

6.5×109 m/sec26.5 \times 10 ^ { 9 } \mathrm {~m} / \mathrm { sec } ^ { 2 }

Answer

6.5×1015 m/sec26.5 \times 10 ^ { 15 } \mathrm {~m} / \mathrm { sec } ^ { 2 }

Explanation

Solution

F = ma = qvB ⇒ a=qvBm=1.6×1019×2×3.4×1071.67×1027a = \frac { q v B } { m } = \frac { 1.6 \times 10 ^ { - 19 } \times 2 \times 3.4 \times 10 ^ { 7 } } { 1.67 \times 10 ^ { - 27 } }

= 6.5 × 1015m/sec2