ℓ=10×pitch
=10×vcos60∘×gB2πm
ℓ=qB10πmv
Put in the value of given data we find ℓ=0.44m
JEE Main 2020 — Physics Electromagnetism
The figure shows a region of length 'l' with a uniform magnetic field of 0.3T in it and a proton entering the region with velocity 4×105ms−1 making an angle 60∘ with the field. If the proton completes 10 revolution by the time it cross the region shown, 'l' is close to (mass of proton =1.67×10−27kg, charge of the proton=1.6×10−19C)

Held on 2 Sept 2020 · Verified 6 Jul 2026.
0.11m
0.88m
0.44m
0.22m
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