## Related questions with answers

Particle accelerators, such a s the Large Hadron Collider, use magnetic fields to steer charged particles a round a ring. Consider a proton ring with 36 identical bending mag nets connected by straight segments. The protons move along a 1.0-m-long circular arc as they pass through each magnet. What magnetic field strength is needed i n each magnet to steer protons around the ring with a speed of $2.5 \times 10^{7} \mathrm{m} / \mathrm{s}$? Assume that the field is uniform inside the magnet, zero outside.

Solutions

VerifiedThe cyclotron motion is the moving of a particle perpendicular to the magnetic field in a uniform circular motion with constant speed. The circular motion is produced by the magnetic field is related to the radius of the motion by equation (29.20) in the form

$\begin{equation} r_{\mathrm{cyc}}=\frac{m v}{q B} \end{equation}$

Where $q$ is the charge of the particle, $v$ is the speed of the particle, $B$ is the magnetic field and $m$ is the mass of the particles. The length of one magnet is 1 m, so for a circular path of 36 magnets, the length of the path will be $L$ = 1 m $\times 36$ = 36 m which represents the circumference of the path. So, we can use it to get the radius of the motion by

$\begin{gather*} L = 2 \pi r_{\mathrm{cyc}}\\ r_{\mathrm{cyc}} = \frac{L}{2\pi} = \frac{36 \mathrm{~m}}{2\pi}\\ r_{\mathrm{cyc}} = 5.73 \mathrm{~m} \end{gather*}$

Our target is to find the required magnetic field, so we rearrange equation (1) for $B$ to be

$\begin{equation} B=\frac{m v}{q r_{\mathrm{cyc}}} \end{equation}$

Now, we plug the values for $m,v,q$ and $r_{\mathrm{cyc}}$ into equation (2) tp get $B$ by

$\begin{align*} B&=\frac{m v}{q r_{\mathrm{cyc}}}\\ &= \frac{\left(1.67 \times 10^{-27} \mathrm{~kg}\right)\left(2.5 \times 10^{7} \mathrm{~m} / \mathrm{s}\right)}{(5.73 \mathrm{~m})\left(1.6 \times 10^{-19} \mathrm{~C}\right)}\\ &= 45.5 \times 10^{-3} \mathrm{~T}\\ &= \boxed{45.5 \mathrm{~mT}} \end{align*}$

$\textbf{Given values:}$

$m=1.67 \times 10^{-27} \: \text{kg}$

$q=1.6 \times 10^{-19} \: \text{C}$

$v=2.5 \times 10^{7} \: m/s$

$c=36$

From $F_{cen}=F_{mag}$, we find $B$:

$\begin{align*} F_{c}&=F_{m}\\ \frac{mv^2}{r}&=qvB\\ \rightarrow \frac{mv}{r}&=qB\\ \rightarrow B&=\frac{mv}{qr} \tag{equation 1.}\\ \end{align*}$

To solve for B, first we have to find radius $r$.

$\begin{align*} c=36 \times 1 \: \text{m}&=2 \pi r\\ 36 \: \text{m}&=2 \pi r\\ \rightarrow r&=\frac{36}{2 \pi}\\ \rightarrow r&=\frac{18}{\pi}\\ \rightarrow r&=5.73 \: \text{m}\\ \end{align*}$

Finally, substitute values in equation 1 and solve for $B$:

$\begin{align*} B&=\frac{mv}{qr}\\ B&=\frac{(1.67 \times 10^{-27} \: \text{kg}) \times (2.5 \times 10^7 \: m/s)}{(1.6 \times 10^{-19} \: \text{C} \times 5.73 \: \text{m})}\\ B&=0.04554 \: \text{T}\\ \end{align*}$

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