Question

1. a proton enters a uniform magnetic field of strength 0.50 t with a speed of 4.0 × 10⁶ m/s, perpendicular to the field. calculate the diameter of the circular path.
2. how would the radius change if:

a) the speed of the particle is doubled? b) the magnetic field strength is doubled?

Explanation:

Step1: Recall radius formula

For a charged particle perpendicular to a magnetic field, the centripetal force is provided by the magnetic force:
$$r = \frac{mv}{qB}$$
where $m=1.67\times10^{-27}\ \text{kg}$ (proton mass), $q=1.6\times10^{-19}\ \text{C}$ (proton charge), $v=4.0\times10^6\ \text{m/s}$, $B=0.50\ \text{T}$

Step2: Calculate initial radius

$$r = \frac{(1.67\times10^{-27})(4.0\times10^6)}{(1.6\times10^{-19})(0.50)} = 0.0835\ \text{m}$$

Step3: Find diameter (Problem 9)

Diameter $d=2r$
$$d = 2\times0.0835 = 0.17\ \text{m}$$

Step4: Analyze speed doubling (10a)

From $r=\frac{mv}{qB}$, $r\propto v$. If $v'=2v$,
$$r' = \frac{m(2v)}{qB} = 2r$$
Radius doubles.

Step5: Analyze B doubling (10b)

From $r=\frac{mv}{qB}$, $r\propto \frac{1}{B}$. If $B'=2B$,
$$r' = \frac{mv}{q(2B)} = \frac{r}{2}$$
Radius is halved.

Answer:

1. $0.17\ \text{m}$
2. a) The radius doubles (becomes $0.17\ \text{m}$)

b) The radius is halved (becomes $0.042\ \text{m}$)