Magnetic Effects & Magnetism

JEE Main level, numerical answer, medium

131 of 276

Question 131 of 276

A proton with a kinetic energy of 2.0eV2.0eV moves into a region of uniform magnetic field of magnitude π2×10−3T\dfrac{\pi }{2}\times {10}^{-3}T. The angle between the direction of magnetic field and velocity of proton is 60o{60}^{o}. The pitch of the helical path taken by the proton is ____ cmcm. (Take, mass of proton =1.6×10−27kg=1.6\times {10}^{-27}kg and charge on proton =1.6×10−19C=1.6\times {10}^{-19}C).

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Answer: 40

The given data is
K.E=2eVK.E=2eV
B=π2×10−3TB=\dfrac{\pi }{2}\times {10}^{-3}T
θ=60o\theta ={60}^{o}
The pitch of the proton is given by
P=2πmqB×vcos⁡θ...(i)P=\dfrac{2\pi m}{qB}\times v\cos \theta ...(i)
By using
K=mv22⇒v=2KmK=\dfrac{m{v}^{2}}{2} \Rightarrow v=\sqrt{\dfrac{2K}{m}}
Substituting the values in equation (i)
=2π×2mKE×12×21.6×10−19×π×10−3=\dfrac{2\pi \times \sqrt{2mKE}\times \dfrac{1}{2}\times 2}{1.6\times {10}^{-19}\times \pi \times {10}^{-3}}
=2×2×1.6×10−27×2×1.6×10−19×1031.6×10−19=\dfrac{2\times \sqrt{2\times 1.6\times {10}^{-27}\times 2\times 1.6\times {10}^{-19}}\times {10}^{3}}{1.6\times {10}^{-19}}
=2×2×10–1=0.4m=2\times 2\times {10}^{–1}=0.4m

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