Question

An electron moves in a circular path perpendicular to a magnetic field of magnitude 0.275 T. If the kinetic energy of the electron is 4.90 10-19 J, find the speed of the electron and the radius of the circular path.

Answer 1

Answer:

The speed of the electron and the radius of the circular path are $10.38\times10^{5}\ m/s$ and $2.15\times10^{-5}\ m$.

Explanation:

Given that,

Magnetic field = 0.275 T

Kinetic energy $K.E= 4.90\times10^{-19}\ J$

We need to calculate the speed of the electron

Using kinetic energy

$K.E=\dfrac{1}{2}mv^2$

$v=\sqrt{\dfrac{2K.E}{m}}$

Where, m = mass of electron

K.E = kinetic energy

Put the value into the formula

$v= \sqrt{\dfrac{2\times4.90\times10^{-19}}{9.1\times10^{-31}}}$

$v=10.38\times10^{5}\ m/s$

We need to calculate the radius of the circular path

Using equation of force on charge in magnetic field

$F = qvB\sin\theta$....(I)

Using centripetal force

$F=\dfrac{mv^2}{r}$....(II)

From equation (I) and (II)

$\dfrac{mv^2}{r}=qvB\sin\theta$

$r=\dfrac{mv^2}{qvB\sin\theta}$

Put the value into the formula

$r=\dfrac{9.1\times10^{-31}\times10.38\times10^{5}}{1.6\times10^{-19}\times0.275\sin90^{\circ}}$

$r=\dfrac{9.1\times10^{-31}\times10.38\times10^{5}}{1.6\times10^{-19}\times0.275}$

$r=2.15\times10^{-5}\ m$

Hence, The speed of the electron and the radius of the circular path are $10.38\times10^{5}\ m/s$ and $2.15\times10^{-5}\ m$.

Answer 2

Answer:

Radius, $r=2.14\times 10^{-5}\ m$

Explanation:

It is given that,

Magnetic field, B = 0.275 T

Kinetic energy of the electron, $E=4.9\times 10^{-19}\ J$

Kinetic energy is given by :

$E=\dfrac{1}{2}mv^2$

$v=\sqrt{\dfrac{2E}{m}}$

$v=\sqrt{\dfrac{2\times 4.9\times 10^{-19}}{9.1\times 10^{-31}}}$            

v = 1037749.04 m/s

The centripetal force is balanced by the magnetic force as :

$qvB\ sin90=\dfrac{mv^2}{r}$

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

$r=\dfrac{9.1\times 10^{-31}\times 1037749.04}{1.6\times 10^{-19}\times 0.275 }$

$r=2.14\times 10^{-5}\ m$

So, the radius of the circular path is $2.14\times 10^{-5}\ m$. Hence, this is the required solution.