A charge is accelerated from rest through a potential difference V and then enters a uniform magnetic field oriented perpendicular to its path. The field deflects the particle into a circular arc of radius R = 5 cm.

(a) If the accelerating potential is tripled to 3V, what will be the radius of the circular arc?

(b) What is the angular frequency omega of rotation, if the mass of this particle is m = 2.0 * 10^-12 kg, its charge is q = 5 mu C, and its initial velocity is v_0 = 5 * 10^4 m/s? (Assume this velocity is a result of the tripled potential, so its radius is that from part (a))

Question

Andrew Chen

Physics

24 March, 21:28

A charge is accelerated from rest through a potential difference V and then enters a uniform magnetic field oriented perpendicular to its path. The field deflects the particle into a circular arc of radius R = 5 cm.

(a) If the accelerating potential is tripled to 3V, what will be the radius of the circular arc?

(b) What is the angular frequency omega of rotation, if the mass of this particle is m = 2.0 * 10^-12 kg, its charge is q = 5 mu C, and its initial velocity is v_0 = 5 * 10^4 m/s? (Assume this velocity is a result of the tripled potential, so its radius is that from part (a))

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Answers (1)

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Juliana Owens · Answer 1

If V be the potential difference which creates velocity v

Ve = 1/2 m v²

e is charge on the particle, m be the mass.

If V is tripled, v becomes √3 times, so velocity becomes √3 v.

in magnetic field, the path becomes circular

for radius of the circle, the relation is

m v² / R = Be v, B is magnetic field

R = m v / B e

So R is proportional to velocity

If velocity becomes √3 times, radius too becomes √3 times

new radius = √3 x 5

= 8.66 cm.

b)

For angular frequency ω = v / r = velocity / radius

velocity = 5 x 10⁴ m / s, r = 8.66 x 10⁻² m

= 5 x 10⁴ / 8.66 x 10⁻²

=.577 x 10⁶ rad / s

= 5.77 x 10⁵ rad / s