Question

A charged particle is accelerated through a potential difference of 12 kV and acquires a speed of 1.0 × 106 m s−1. It is then injected perpendicularly into a magnetic field of strength 0.2 T. Find the radius of the circle described by it.

Answer 1

The radius of the circle is 12 cm.

Explanation:

F(m) = qvB sin(90°)

Applied potential difference "V" = 12 kV = 12 × 10^3 V

Speed of a charged particle, v =1.0 × 10^6 m s−1

Magnetic field strength "B" = 0.2 T

As per the question, a charged particle is injected perpendicularly into the magnetic field.

We know:

qV = 1 / 2mv^2

m / q = 2V / v^2

= 2 × 12×10^3  / (1×106)^2

= 24×10^−9

r = mv / qB

r=24×10^−9 × 10^6 / 0.2

r = 12 × 10^−2 m

= 12 cm

Thus the radius of the circle is 12 cm.

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