# A magnetic field of 0.3T is present in a region and a proton enters the region with a velocity 4 x 105 m/s making an angle of 600 with the field. If the proton completes 10 revolutions by the time it crosses the region as shown in the figure, then $l$ is close to (mass of proton=1.67 x 10-27 kg, a charge of proton=1.6 x 10-19 C, Answer in meter ) Option: 1 0.44 Option: 2 0.88 Option: 3 0.22 Option: 4 0.11

Given

m = 1.67 x 10-27 kg

v = 4 x 105 m/s

q = 1.6 x 10-19 C

B = 0.3 T

Using

$T= \frac{2 \pi m}{q B}$ and

$pitch=v \cos 60^{\circ} \times T=v \cos 60^{\circ} \times \frac{2 \pi m}{q B}$

we get

\begin{aligned} l &=10 \times \text { pitch } \\ &=10 \times v \cos 60^{\circ} \times \frac{2 \pi m}{q B} \\ &=10 \times v \cos 60^{\circ} \times \frac{2 \pi m}{q B} \\ &=10 \times v \times \frac{1}{2} \times \frac{2 \pi m}{q B} \\ l &=\frac{10 v \pi m}{q B}=\frac{10 \times 4 \times 10^{5} \times 3.14 \times 1.67 \times 10^{-27}}{1.6 \times 10^{-19} \times 0.3} \cong & 0.44 m \end{aligned}

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