Magnetic Effects & Magnetism

JEE Main level, single correct, medium

152 of 276

Question 152 of 276

The figure shows a region of length 'll' with a uniform magnetic field of 0.3T0.3T in it and a proton entering the region with velocity 4×105ms−14\times {10}^{5}m{s}^{-1} making an angle 60∘60^{\circ} with the field. If the proton completes 1010 revolution by the time it cross the region shown, 'll' is close to (mass of proton =1.67×10−27kg=1.67\times {10}^{-27}kg, charge of the proton=1.6×10−19C=1.6\times {10}^{-19}C)

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Answer: (C) 0.44m0.44m

ℓ=10×pitch\ell =10\times pitch
=10×vcos⁡60∘×2πmgB=10\times v\cos 60^{\circ}\times \dfrac{2\pi m}{gB}
ℓ=10πmvqB\ell =\dfrac{10\pi mv}{qB}
Put in the value of given data we find ℓ=0.44m\ell =0.44m

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