KCET · Physics · Magnetic Effects of Current
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
- A \( 0.1 \mathrm{~m} \)
- B \( 10 \mathrm{~m} \)
- C \( 100 \mathrm{~m} \)
- D \( 1 \mathrm{~m} \)
Answer & Solution
Correct Answer
(C) \( 100 \mathrm{~m} \)
Step-by-step Solution
Detailed explanation
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 \( 100 \mathrm{~m} \).
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