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Question: One proton beam enters a magnetic field of \(10 ^ { - 4 }\)T normally, Specific charge = \(10 ^ { 1...

One proton beam enters a magnetic field of 10−410 ^ { - 4 }T normally, Specific charge = 1011C/kg10 ^ { 11 } \mathrm { C } / \mathrm { kg } velocity = 107 m/s10 ^ { 7 } \mathrm {~m} / \mathrm { s } What is the radius of the circle described by it

A

0.1 m

B

1 m

C

10 m

D

None of these

Answer

1 m

Explanation

Solution

r=mvqB=1071011×10−4=1m(∵q/m=1011C/kg)r = \frac { m v } { q B } = \frac { 10 ^ { 7 } } { 10 ^ { 11 } \times 10 ^ { - 4 } } = 1 m \left( \because q / m = 10 ^ { 11 } \mathrm { C } / \mathrm { kg } \right)