A proton moves at 5.0 × 10⁶ m/s in a uniform field of 0.30 T…

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Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

A proton moves at $5.0 \times 10^{6}$ m/s in a uniform field of $0.30$ T. Its velocity makes an angle of $20^\circ$ with the field. Find the radius of its helix, the pitch, and the number of turns it makes while it travels $5.0$ m along the field. Use $m = 1.67 \times 10^{-27}$ kg.

Need a reminder? The key ideaShow

At right angles to the field, $r = \dfrac{mv}{qB}$ , $T = \dfrac{2\pi m}{qB}$ and $f = \dfrac{qB}{2\pi m}$ . After a voltage $V$ , $v = \sqrt{2qV/m}$ . At an angle $\theta$ the path is a helix with $r = \dfrac{mv\sin\theta}{qB}$ and pitch $p = \dfrac{2\pi m\,v\cos\theta}{qB}$ . A velocity selector passes $v = E/B$ . In a cyclotron, $f = \dfrac{qB}{2\pi m}$ and $K_{\max} = \dfrac{q^2B^2R^2}{2m}$ .

The worked solution

Have you tried it?

Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

Plan. The part across the field makes the circle. The part along it carries the proton on.

Step 1. The two parts. $v_\perp = v\sin 20^\circ = 5.0 \times 10^{6} \times 0.342 = 1.71 \times 10^{6}\ \text{m/s}$ . $v_\parallel = v\cos 20^\circ = 5.0 \times 10^{6} \times 0.940 = 4.70 \times 10^{6}\ \text{m/s}$ .

Step 2. The radius. The top of $r = \dfrac{mv_\perp}{eB}$ is $mv_\perp = 1.67 \times 10^{-27}\ \text{kg} \times 1.71 \times 10^{6}\ \text{m/s} = 2.86 \times 10^{-21}\ \text{kg m/s}$ . The bottom is $eB = 1.6 \times 10^{-19}\ \text{C} \times 0.30\ \text{T} = 4.8 \times 10^{-20}\ \text{C T}$ . So $r = \dfrac{2.86 \times 10^{-21}\ \text{kg m/s}}{4.8 \times 10^{-20}\ \text{C T}} = 5.95 \times 10^{-2}\ \text{m}$ , that is $5.95$ cm.

Step 3. The period. $T = \dfrac{2\pi m}{eB} = \dfrac{2\pi \times 1.67 \times 10^{-27}\ \text{kg}}{4.8 \times 10^{-20}\ \text{C T}} = 2.19 \times 10^{-7}\ \text{s}$ .

Step 4. The pitch. $p = v_\parallel T = 4.70 \times 10^{6}\ \text{m/s} \times 2.19 \times 10^{-7}\ \text{s} = 1.03\ \text{m}$ .

Step 5. The turns. $N = \dfrac{5.0\ \text{m}}{1.03\ \text{m}} = 4.87$ .

Answer: $5.95$ cm, $1.03$ m and $4.87$ turns.

Check. The helix is long and thin: the radius is $6$ cm and the pitch is $103$ cm. That is right for a velocity that is only $20^\circ$ from the field.

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