Question

An electron is accelerated from rest through a potential difference of 3750 V. It enters a region where the magnetic field is 4.0 mT perpendicular to the velocity of the electron. Calculate the radius of the path this electron will follow.

Answer 1

Answer:

r=5.1 cm

Explanation:

Given that

Potential difference ΔV= 3750 V

Magnetic filed B= 4 m T

When electron moves in circular track

The radial force Fr

$Fr=\dfrac{mv^2}{r}$

The force due to magnetic filed

F= e v B

So we can say that

F= Fr

$q v B=\dfrac{mv^2}{r}$

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

Now from energy conservation

$e\Delta V=\dfrac{1}{2}mv^2$

We know that

The mass of electron

$m=9.1\times 10^{-31}\ kg$

The charge on electron

$q=e=1.6\times 10^{-19}\ C$

$v=\sqrt{\dfrac{2e\Delta V}{m}}$

Now by putting the values

$v=\sqrt{\dfrac{2\times 1.6\times 10^{-19}\times 3750}{9.1\times 10^{-31}}}$

v=36.31 x 10⁶ m/s

Now

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

$r=\dfrac{36.31\times 10^6\times 9.1\times 10^{-31}}{1.6\times 10^{-19}\times 4\times 10^{-3}}$

r=0.051 m

r=5.1 cm

So the radius of path r= 5.1 cm