he speed of the proton is 2.0 × 106 m s–1 and the magnetic field strength B is 0.35 T.(i) Show that the radius of the path is about 6 cm.
Question
he speed of the proton is 2.0 × 10^6 m s–1 and the magnetic field strength B is 0.35 T.
(i) Show that the radius of the path is about 6 cm.
Solution
To calculate the radius of the path of a charged particle moving in a magnetic field, we can use the formula:
r = mv / (qB)
where:
- r is the radius of the path,
- m is the mass of the particle,
- v is the speed of the particle,
- q is the charge of the particle, and
- B is the magnetic field strength.
Given in the problem, we have:
- v = 2.0 × 10^6 m/s,
- B = 0.35 T,
- the charge of a proton q = 1.6 × 10^-19 C, and
- the mass of a proton m = 1.67 × 10^-27 kg.
Substituting these values into the formula, we get:
r = (1.67 × 10^-27 kg * 2.0 × 10^6 m/s) / (1.6 × 10^-19 C * 0.35 T)
Solving this equation gives r ≈ 0.06 m, or about 6 cm. Therefore, the radius of the path is about 6 cm.
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