Stacie Worsley

2021-12-16

An electron is accelerated from rest by a potential difference of 350 V. It then enters a uniform magnetic field of magnitude 200 mT with its velocity perpendicular to the field. Calculate (a) the speed of the electron and (b) the radius of its path in the magnetic field.

Mary Herrera

Beginner2021-12-17Added 37 answers

a) Assume we have an electron accelerated using potential difference of $V=350\text{}V$ , which gives the ion a speed of v, to find this speed we set the potential energy equals to the kinetic energy of the ion:

$eV=\frac{1}{2}m{v}^{2}$

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

substitute with the givens to get:

$v=\sqrt{\frac{2(1.6\times {10}^{-19}\text{}C)\left(350\text{}V\right)}{9.11\times {10}^{-31}\text{}kg}}$

$=1.11\times {10}^{7}\text{}\frac{m}{s}$

b) Then the electron enters a region of uniform magnetic field, it moves in circular path with radius of:

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

substitute with givens to get:

$r=\frac{(9.11\times {10}^{-31}\text{}kg)(1.11\times {10}^{7}\text{}\frac{m}{s})}{(1.6\times {10}^{-19}\text{}C)(200\times {10}^{-3}\text{}T)}$

$=3.16\times {10}^{-4}\text{}m$

$r=3.16\times {10}^{-4}\text{}m$

Result: a)$v=1.11\times {10}^{7}\text{}\frac{m}{s}$

b)$r=3.16\times {10}^{-4}\text{}m$

substitute with the givens to get:

b) Then the electron enters a region of uniform magnetic field, it moves in circular path with radius of:

substitute with givens to get:

Result: a)

b)

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