A beam of protons with a velocity of 4×105 ms−1  enters a uniform magnetic field of 0.3T.  The velocity makes an angle of  60° with the magnetic field. Find the radius of the helical path taken by the proton beam

# A beam of protons with a velocity of $4×{10}^{5}\text{\hspace{0.17em}}{\mathrm{ms}}^{-1}$  enters a uniform magnetic field of $0.3\mathrm{T}.$  The velocity makes an angle of  60° with the magnetic field. Find the radius of the helical path taken by the proton beam

1. A

1.2 cm

2. B

1.6cm

3. C

1.7cm

4. D

1.8cm

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### Solution:

The components of the proton’s velocity along and perpendicular to the magnetic field are ${\mathrm{v}}_{\parallel }=\left(4×{10}^{5}{\mathrm{ms}}^{-1}\right)\mathrm{cos}{60}^{0}=2×{10}^{5}{\mathrm{ms}}^{-1}.$

And   ${\mathrm{v}}_{\perp }=\left(4×{10}^{5}{\mathrm{ms}}^{-1}\right)\mathrm{sin}{60}^{\circ }=2\sqrt{3}×{10}^{5}{\mathrm{ms}}^{-1}$
As the force $\mathrm{q}\stackrel{\to }{\mathrm{v}}×\stackrel{\to }{\mathrm{B}}$ is perpendicular to the magnetic field, the component ${\mathrm{v}}_{\parallel }$ will remain constant. In the plane perpendicular to the field, the proton will describe a circle whose radius is obtained from the equation ${\mathrm{qv}}_{\perp }\mathrm{B}=\frac{{\mathrm{mv}}_{\perp }^{2}}{\mathrm{r}}$
Or   $\mathrm{r}=\frac{{\mathrm{mv}}_{\perp }}{\mathrm{qB}}=\frac{\left(1.67×{10}^{-27}\mathrm{kg}\right)×\left(2\sqrt{3}×{10}^{5}{\mathrm{ms}}^{-1}\right)}{\left(1.6×{10}^{-19}\mathrm{C}\right)×\left(0.3\mathrm{T}\right)}=0.012\mathrm{m}=1.2\mathrm{cm}$

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