A uniform magnetic field of 0.50 T is directed along the positive x axis. A proton moving with a speed of 60 km s enters this fi

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A uniform magnetic field of 0.50 T is directed along the positive x axis. A proton moving with a speed of 60 km s enters this fi

eld. The helical path followed by the proton shown has a pitch of 5.0 mm. Determine the angle between the magnetic field and the velocity of the proton.

Physics

1 answer:

Keith_Richards [3.2K] · Answer 1 · 2026-03-24T18:08:00+03:00

Given values are as follows:

Magnetic field, B = 0.5 T

Proton's velocity, v = 60 km/s = 60000 m/s

The helix traced by the proton has a pitch of 5.0 mm, p = 0.005 m.

Our goal is to ascertain the angle formed between the magnetic field and the proton's velocity. The pitch of the helix equates to the product of the parallel velocity component and the time period. In mathematical terms, this is denoted as:

$p=v_{||}\times T$

$p=v\ cos\theta\times \dfrac{2\pi m}{Bq}$

$cos\theta=\dfrac{pBq}{2\pi mv}$

$cos\theta=\dfrac{0.005\times 0.5\times 1.6\times 10^{-19}}{2\pi \times 1.67\times 10^{-27}\times 60000}$

$\theta=50.58^{\circ}$

Thus, the angle between the magnetic field and the proton's velocity is determined to be 50.58 degrees. This concludes the solution.