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प्रश्न
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
उत्तर
Given:
Mass of the proton, m = 1.6 × 10−27 kg
Speed of the proton inside the crossed electric and magnetic field, v = 2.0 × 105 ms−1
As per the question, the proton is not deflected under the combined action of the electric and magnetic fields. Thus, the forces applied by both the fields are equal and opposite.
That is, qE = qvB
⇒ E = vB ...(1)
But when the electric field is switched off, the proton moves in a circle due to the force of the magnetic field.
Radius of the circle, r = 4.0 cm = 4 × 10−2 m
We know: r = `(mv)/(qB)`
⇒ `(1.6xx10^-27xx2xx10^5)/(1.6xx100^194xx10^-3`)
= 0.5 × 10−1 = 0.05 T
Putting the value of B in equation (1), we get:
E = 2 × 105 × 0.05
= 1 × 104 N/c
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