Question

A proton projected in a magnetic field of 0.020 T travels along a helical path of radius 5.0 cm and pitch 20 cm. Find the components of the velocity of the proton along and perpendicular to the magnetic field. Take the mass of the proton = 1.6 × 10−27 kg

Answer 1

Answer:

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Answer 2

Explanation:

Proton’s mass, mp = $1.6 \times 10^{-27} \mathrm{kg}$

Intensity of the magnetic field, $B=0.02 \mathrm{T}$

Helical path’s radius, $r=5 \times 10^{-2} \mathrm{m}$

Helical path’s pitch, $p=2 \times 10^{-1} \mathrm{m}$

It is understood that for a helical path, the proton has two components, v∥ and v⊥.

Now,

$m v \perp 2 r=q v \perp B$

$\Rightarrow r=m v \perp q B \Rightarrow 5 \times 10^{-2}$

$\Rightarrow v \perp=105 \mathrm{m} / \mathrm{s}$

$\text { Pitch }=v\|2 \pi r v \perp v\|=0.6369 \times 105=6.4 \times 104 \mathrm{m} / \mathrm{s}$