Question

An electron moving at right angle to a uniform magnetic field comlete a circulat orbit in 1 microsecond find the magnetic field

Answer 1

Answer: $\textbf{35.7 \mu T}$


Let the charge on electron be $q$ and mass be $m$. Also, let the Magnetic Field be $B$

Also, let us assume that the velocity of the electron as it moves through the magnetic field is $v$.


Now, Force on an electron moving through Magnetic Field is given by:

$\vec{F} = q \, \, \vec{v} \times \vec{B}$

Here, the electron is moving at right angle to magnetic field. So, Force is:


$F = q v B \sin 90^{\circ} = qvB$


This Force is the Centripetal Force which keeps the electron revolving in a circular path. Let us say that the radius of the Circular Path is $R$

Then

$F = \frac{mv^2}{R} \\ \\ \\ \implies qvB = \frac{mv^2}{R} \\ \\ \\ \implies B = \frac{mv}{qR}$

Now, let us say that the angular velocity of the electron is $\omega$

We are given that the electron complete one orbit in 1 microsecond. Since 1 complete circular orbit consists of $2\pi$ radians, we can find the Angular Velocity:

$\omega = \frac{2\pi}{T} \\ \\ \\ \implies \omega = \frac{2\pi}{10^{-6} } \\ \\ \\ \implies \omega = 2\pi \times 10^6 \, \, rad/s$

Also, the relation between velocity and angular velocity is given as:

$\omega = \frac{v}{R}$

So, we can write:

$B = \frac{mv}{qR} \\ \\ \\ \implies B = \frac{m\omega}{q} \\ \\ \\ \implies B = \frac{9.1\times 10^{-31} \times 2\pi \times 10^6}{1.6 \times 10^{-19}} \\ \\ \\ \implies B \approx 3.57 \times 10^{-5} \, \, T \\ \\ \\ \implies \boxed{B \approx 35.7 \, \, \mu T}$

Thus, the Magnetic Field is approximately 35.7 microteslas

Answer 2

ANSWER: 3.57 × 10*-5

EXPLANATION:

T=10*-6 seconds

T=2πr/v v is velocity and T is time period and r is radius

v=Ber/m

so,

T=2πrm/Ber

T=2πm/Be

B=2πm/Te

B=2×3.14×9.1×10*-31/1×10*-6×1.6×10*-19

B=35.7×10*-6

B=3.57×10*-5