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GUJCET · Physics · Moving Charges and Magnetism
An electron is moving at a speed of \(3.2 \times 10^7 ms^{-1}\) in a magnetic field of \(12 \times 10^{-4} T\) perpendicular to the direction of motion of electron. The radius of the path of the electron is _________ \(cm .\left(e=1.6 \times 10^{-19} C\right.\) and \(\left.m_e=9 \times 10^{-31} kg\right)\)
- A 30
- B 13
- C 15
- D 26
Answer & Solution
Correct Answer
(C) 15
Step-by-step Solution
Detailed explanation
C
The radius of the circular path,
\(\begin{array}{l}r=\frac{m U }{q B} \\ r=\frac{9 \times 10^{-31} \times 3.2 \times 10^7}{1.6 \times 10^{-19} \times 12 \times 10^{-4}} \\ r=15 cm\end{array}\)
The radius of the circular path,
\(\begin{array}{l}r=\frac{m U }{q B} \\ r=\frac{9 \times 10^{-31} \times 3.2 \times 10^7}{1.6 \times 10^{-19} \times 12 \times 10^{-4}} \\ r=15 cm\end{array}\)
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