Pitch
=(Vcosθ)T
=(Vcosθ)eB2πm
=(4×105cos60∘)0.3×102π(1.69×10191.67×10−27)
4cm
JEE Main 2020 — Physics Electromagnetism
A beam of protons with speed 4×105ms−1 enters a uniform magnetic field of 0.3T at an angle 60∘ to the magnetic field, the pitch of the resulting helical path of protons is close to : (Mass of the proton=1.67×10−27kg, charge of the proton=1.69×10−19C)
Held on 2 Sept 2020 · Verified 6 Jul 2026.
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