A proton moving with a velocity \(3 \times 10^5 \mathrm{~m} / \mathrm{s}\) enters a magnetic field of \(0.3\) tesla at a…
Q1
A proton moving with a velocity \(3 \times 10^5 \mathrm{~m} / \mathrm{s}\) enters a magnetic field of \(0.3\) tesla at an angle of \(30^{\circ}\) with the field. The radius of curvature of its path will be (e/m for proton \(=10^8 \mathrm{C} / \mathrm{kg}\) )
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