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Question: A proton moving with a velocity, \(2.5 \times 10 ^ { 7 } \mathrm {~m} / \mathrm { s }\), enters a ma...

A proton moving with a velocity, 2.5×107 m/s2.5 \times 10 ^ { 7 } \mathrm {~m} / \mathrm { s }, enters a magnetic field of intensity 2.5T making an angle 30∘30 ^ { \circ } with the magnetic field. The force on the proton is

A
B
C

6×10−126 \times 10 ^ { - 12 }N

D

9×10−12N9 \times 10 ^ { - 12 } N

Answer
Explanation

Solution

F=1.6×10−19×6.25×107×12=5×10−12 NF = 1.6 \times 10 ^ { - 19 } \times 6.25 \times 10 ^ { 7 } \times \frac { 1 } { 2 } = 5 \times 10 ^ { - 12 } \mathrm {~N}