Answer to Question #172692 in Electricity and Magnetism for Todd Phillips

Question

An electron (m = 9.11 x 10^-31kg, q = 1.60 x 10^-19C) at rest is accelerated through an electric potential difference of 12.0 V. It then enters a uniform magnetic field of 0.100 T [up], when initially moving [N].

a) Find the initial speed of the electron entering the magnetic field.

b) Find the magnetic force on the electron entering the magnetic field.

c) Find the radius of the circular path of the electron in the field. 

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 electron accelerates from rest, $v_1=0$. Then, we get:

$$v_2=\sqrt{\dfrac{2q(V_1-V_2)}{m}},$$ $$v_2=\sqrt{\dfrac{2\cdot1.6\cdot10^{-19}\ C\cdot12\ V}{9.11\cdot10^{-31}\ kg}}=2.05\cdot10^6\ \dfrac{m}{s}.$$

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

$$F_B=qvBsin\theta,$$ $$F_B=1.6\cdot10^{-19}\ C\cdot2.05\cdot10^6\ \dfrac{m}{s}\cdot0.1\ T\cdot sin90^{\circ}=3.28\cdot10^{-14}\ N.$$

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

$$r=\dfrac{mv}{qB}=\dfrac{9.11\cdot10^{-31}\ kg\cdot2.05\cdot10^6\ \dfrac{m}{s}}{1.6\cdot10^{-19}\ C\cdot0.1\ T}=1.17\cdot10^{-4}\ m.$$