The 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.

To solve this problem, we need to use the formula for the radius of the path of a charged particle moving in a magnetic field, which is given by:

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 indeed about 6 cm.