Question

A proton goes undeflected in a crossed electric and magnetic field (the fields are perpendicular to each other) at a speed of 2.0 × 105 m s−1. The velocity is perpendicular to both the fields. When the electric field is switched off, the proton moves along a circle of radius 4.0 cm. Find the magnitudes of the electric and magnetic fields. Take the mass of the proton = 1.6 × 10−27 kg

Answer 1

Answer:

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Answer 2

Explanation:

It is given:

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

Inside the crossed magnetic and electric field, the proton speed,$\mathrm{v}=2.0 \times 10^{5} \mathrm{ms}^{-1}$

According to the information provided, the proton is not glanced under the united action of the magnetic and electric fields. So, by both the fields, the applied forces are opposite and equal.

$\Rightarrow \mathrm{E}=\mathrm{vB}$    …(1)

But when there is a switch off of the electric field, there will be a movement of proton in a circular fashion because of the magnetic field’s force.

Circle’s radius, $r=4 \times 10^{-2} \mathrm{m}$

We understand: $r=\mathrm{mvq} \mathrm{B}$

$\Rightarrow B=m v q r=0.5 \times 10^{-1}=0.05 T$

When the value of B is applied in in equation (1), we obtain:

$E=1 \times 104 \mathrm{N} / \mathrm{c}$.