Checkout JEE MAINS 2022 Question Paper Analysis : Checkout JEE MAINS 2022 Question Paper Analysis :

The figure shows a region of length ‘l’ with a uniform magnetic field of 0.3 T in it and a proton entering the region with velocity 4 ×105 ms–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 × 10–27 kg, charge of the proton = 1.6 × 10–19 C)

Solution Paper of JEE Main 2020 Physics Shift2-2nd Sept

1) 0.11 m

2) 0.22 m

3) 0.44 m

4) 0.88 m

Answer: (3)

l = 10 × pitch

= 10 × vcos 60° ×(2π m/qB)

= 10 × v ×(1/2) ×(2πm/qB)

\(\begin{array}{l}l=\frac{10v\pi m}{qB}=\frac{10\times 4\times 10^{5}\times 3.14\times 1.67\times 10^{-27}}{1.6\times 10^{-19}\times 0.3}\simeq 0.44 m\simeq\end{array} \)

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