Q149 / 180 · KCET 2015 · Physics
A proton beam enters a magnetic field of \( 10^{-4} \mathrm{~Wb} m^{-2} \) normally. If the specific charge of the
proton is \( 10^{11} \mathrm{C} \mathrm{kg}^{-1} \) and its velocity is \( 10^{9} \mathrm{~ms}^{-1} \), then the radius of the circle described will
be
proton is \( 10^{11} \mathrm{C} \mathrm{kg}^{-1} \) and its velocity is \( 10^{9} \mathrm{~ms}^{-1} \), then the radius of the circle described will
be
- A\( 0.1 \mathrm{~m} \)
- B\( 10 \mathrm{~m} \)
- Correct answerC\( 100 \mathrm{~m} \)
- D\( 1 \mathrm{~m} \)
Solution
Magnetic field \( B=10^{-4} \mathrm{~W} \mathrm{bm}^{-2} \), charge \( q=10^{11} \mathrm{Ck} g^{-1} \)
velocity \( v=10^{9} \mathrm{~ms}^{-1} \)
Now, radius of the circle described by proton is
\( r=\frac{m v}{q B}=\frac{v}{\left(\frac{q}{m}\right) \times B}=\frac{10^{9}}{10^{11} \times 10^{-4}}=10^{2} \)
Therefore, radius of circle is \( 100 \mathrm{~m} \).
velocity \( v=10^{9} \mathrm{~ms}^{-1} \)
Now, radius of the circle described by proton is
\( r=\frac{m v}{q B}=\frac{v}{\left(\frac{q}{m}\right) \times B}=\frac{10^{9}}{10^{11} \times 10^{-4}}=10^{2} \)
Therefore, radius of circle is \( 100 \mathrm{~m} \).
Hence, the correct answer is option: 'C'.