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A charged particle is accelerated through a potential difference of 12 kV and acquires a speed of 1.0 × 10^6 m/s. It is then injected perpendicularly into a magnetic field of strength 0.2 T. Find the radius of the circle described by it. 35. Doubly-ionized helium ions are projected with a speed of 10 km/s in a direction perpendicular to a uniform magnetic field of magnitude 1.0 T. (a) Find the force acting on an ion, (b) The radius of the circle in which it circulates and (c) the time taken by an ion to complete the circle.

A charged particle is accelerated through a potential difference of 12 kV and acquires a speed of 1.0 × 10^6 m/s. It is then injected perpendicularly into a magnetic field of strength 0.2 T. Find the radius of the circle described by it.
35. Doubly-ionized helium ions are projected with a speed of 10 km/s in a direction perpendicular to a uniform magnetic field of magnitude 1.0 T. (a) Find the force acting on an ion, (b) The radius of the circle in which it circulates and (c) the time taken by an ion to complete the circle.

Grade:10

2 Answers

Jitender Pal
askIITians Faculty 365 Points
7 years ago
Sol. V = 12 KV E = V/l Now, F = qE = qV/ld or, a = F/m = qV/ml υ = 1 * 10^6 m/s Or V = √2 * qV/ml*l = √2 * q/m * 12 * 10^3 Or 1 * 10^6 = √2 * q/m * 12 * 10^3 ⇒ 10^12 = 24 * 10^3 * q/m ⇒ m/q = 24 * 10^3/10^12 = 24 * 10^-9 r = mV/qB = 24 * 10^-9 *1 * 10^6/2 * 10^-1 = 12 * 10^-2 m = 12 cm
Prateek raj
13 Points
3 years ago
Initial ke=final ke
qv=1/2 mv^2
m/q=2V/v^2------eqn-(1)
We know 
R=mv/qB
R=2V/v^2×v/B
R=2V/VB
R=2×12×10^3/10^6×0.2
R=12×10^-2
R=12cm

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