Question

An α-particle has a charge of +2e and a mass of 6.64 × 10^-27 kg. It is accelerated from rest through a potential difference that has a value of 1.85 × 10⁶ V and then enters a uniform magnetic field whose magnitude is 2.65 T. The α-particle moves perpendicular to the magnetic field at all times. What is (a) the speed of the α-particle, (b) the magnitude of the magnetic force on it, and (c) the radius of its circular path?

Answer 1

(a) Let's write the law of conservation of energy:

$$KE_1+PE_1=KE_2+PE_2,$$ $$\dfrac{1}{2}mv_1^2+qV_1=\dfrac{1}{2}mv_2^2+qV_2.$$

Since the alpha-particle accelerates from rest, $v_1=0$. Then, we get:

$$v_2=\sqrt{\dfrac{2q(V_1-V_2)}{m}},$$ $$v_2=\sqrt{\dfrac{2\cdot2\cdot1.6\cdot10^{-19}\ C\cdot1.85\cdot10^6\ V}{6.64\cdot10^{-27}\ kg}}=1.33\cdot10^7\ \dfrac{m}{s}.$$

(b) We can find the magnitude of the magnetic force as follows:

$$F_B=qvBsin\theta,$$ $$F_B=2\cdot1.6\cdot10^{-19}\ C\cdot1.33\cdot10^7\ \dfrac{m}{s}\cdot2.65\ T\cdot sin90^{\circ}=1.13\cdot10^{-11}\ N.$$

(c) We can find the radius of the circular path from the formula:

$$r=\dfrac{mv}{qB}=\dfrac{6.64\cdot10^{-27}\ kg\cdot1.33\cdot10^7\ \dfrac{m}{s}}{2\cdot1.6\cdot10^{-19}\ C\cdot2.65\ T}=0.104\ m.$$