Magnetic Effects & Magnetism

JEE Main level, single correct, medium

154 of 276

Question 154 of 276

A beam of protons with speed 4×105ms−14\times {10}^{5}m{s}^{-1} enters a uniform magnetic field of 0.3T0.3T at an angle 60∘60^{\circ} to the magnetic field, the pitch of the resulting helical path of protons is close to : (Mass of the proton=1.67×10−27kg=1.67\times {10}^{-27}kg, charge of the proton=1.69×10−19C=1.69\times {10}^{-19}C)

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Answer: (D) 4cm4cm

Pitch
=(Vcosθ)T=(Vcos\theta )T
=(Vcosθ)2πmeB=(Vcos\theta )\dfrac{2\pi m}{eB}
=(4×105cos⁡60∘)2π0.3×10(1.67×10−271.69×1019)=\left(4\times {10}^{5}\cos 60^{\circ}\right)\dfrac{2\pi }{0.3\times 10}\left(\dfrac{1.67\times {10}^{-27}}{1.69\times {10}^{19}}\right)
4cm4cm

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