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MHT CET · Physics · Magnetic Effects of Current

Two particles A and B have equal charges but different masses \(\mathrm{M}_{\mathrm{A}}\) and \(\mathrm{M}_{\mathrm{B}}\). After being accelerated through same potential difference enter the region of uniform
magnetic field and describe the path of radii \(\mathrm{R}_{\mathrm{A}}\) and \(\mathrm{R}_{\mathrm{B}}\) respectively. Then \(\mathrm{M}_{\mathrm{A}}: \mathrm{M}_{\mathrm{B}}\)
is

  1. A \(\frac{\mathrm{R}_{\mathrm{A}}}{\mathrm{R}_{\mathrm{B}}}\)
  2. B \(\frac{\mathrm{R}_{\mathrm{B}}}{\mathrm{R}_{\mathrm{A}}}\)
  3. C \(\left(\frac{\mathrm{R}_{\mathrm{A}}}{\mathrm{R}_{\mathrm{B}}}\right)^{2}\)
  4. D \(\left(\frac{\mathrm{R}_{\mathrm{B}}}{\mathrm{R}_{\mathrm{A}}}\right)^{2}\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(\left(\frac{\mathrm{R}_{\mathrm{A}}}{\mathrm{R}_{\mathrm{B}}}\right)^{2}\)

Step-by-step Solution

Detailed explanation

\(qV = \frac{1}{2}mv^2 \implies v = \sqrt{\frac{2qV}{m}}\) \(R = \frac{mv}{qB} = \frac{m}{qB} \sqrt{\frac{2qV}{m}} = \frac{1}{qB} \sqrt{2mqV}\)
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