A proton moving with a velocity $ 3 \times 10^5 \, \text{m/s} $ enters a magnetic field of $ 0.3 \, \text{tesla} $ at an angle of $ 30^\circ $ with the field. The radius of curvature of its path will be $ e/m $ for proton $ = 10^8 \, \text{C/kg} $.

Question

Accepted Answer

1.25 cm. The radius of curvature is given by the formula $ r = \frac{mv}{qB \sin \theta} $. Substituting $ m/q = 1/(e/m) = 1/(10^8) $, $ v = 3 \times 10^5 $, $ B = 0.3 $, and $ \sin 30^\circ = 0.5 $, we get $ r = \frac{3 \times 10^5}{10^8 \times 0.3 \times 0.5} = 0.0125 \, \text{m} = 1.25 \, \text{cm} $.

A proton moving with a velocity \( 3 \times 10^5 \, \text{m/s} \) enters a magnetic field of \( 0.3 \, \text{tesla} \) at an angle of \( 30^\circ \) with the field. The radius of curvature of its path will be \( e/m \) for proton \( = 10^8 \, \text{C/kg} \).

Solution

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