Question

2.032 4. An electron is accelerated through potential difference of 150 V from rest and then enters a region of uniform magnetic field traveling perpendicular to the field. The magnitude of magnetic field is 0.50 T. a) What is the magnitude of magnetic force acting on the electron (5.8x10-14 N), and b) what is the radius of the path? (8.3x10 m) Note: The mass of electron is 9.11 x 103 kg, and the charge of electron is 1.6x 10-1 C in absolute value.

Answer 1

Let us first calculate speed of the electron when it get accelerated through 150 V, Using conservation of energy

Let us first calculate speed of the electron when it get accelerated through 150 V, Using conservation of energy 1/2 mv^2 = qv v = squareroot 2 q V/m v = squareroot 2 times 1.6 times 10^- 19 C times 150 V/9.11 times 10^- 31 kg = 7.26 times 10^6 m/s Part A) Magnetic force on the electron inside the magnetic force F = qv B sin theta = 1.6 times 10^- 19 C times 7.26 times 10^6 m/s times 0.50 T times sin 90 degree = 5.80 times 10^- 13 N Part B) Radius of the Path mv^2/r = qvB sin 90 degree mv/r = qB r = mv/qB r = 9.11 times 10^- 31 kg times 7.26 times 10^6 m/s/1.6 times 10^- 19 C times 0.50 T = 8.3 times 10^- 5 m

$v= \sqrt{\frac{2qV}{m}}$

$v= \sqrt{\frac{2\times 1.6\times 10^{-19}C\times 150 V}{9.11\times 10^{-31}kg}}=7.26\times 10^{6} m/s$

Part A) Magnetic force on the electron inside the magnetic force

$F =qvB\sin \theta =1.6\times 10^{-19}C\times 7.26\times 10^{6} m/s \times 0.50 T\times \sin 90^{0}=5.80\times 10^{-13}N$

Part B) Radius of the Path

$\frac{mv^{2}}{r} =qvB\sin 90^{0}$

$\frac{mv}{r} =qB$

$r = \frac{mv}{qB}$

$r = \frac{9.11\times 10^{-31}kg\times 7.26\times 10^{6} m/s}{1.6\times 10^{-19}C\times 0.50T}=8.3\times 10^{-5} m$