Question

An electron moves with a speed of 8.0×106m/s along the -z-axis. It enters a region where there is a uniform magnetic field B = (5.5T)i – (3.7T)j. What is the acceleration of the electron when it first enters the region of the uniform magnetic field?

Answer 1

Answer:

Acceleration, $a=9.36\times 10^{18}\ m/s^2$

Explanation:

It is given that,

Speed of electron, $v=8\times 10^6\ m/s$

Charge on an electron, $q=1.6\times 10^{-19}\ C$

Mass of electron, $m=9.1\times 10^{-31}\ kg$

Magnetic field, $B=5.5i-3.7j$

Magnitude, $|B|=\sqrt{5.5^2+(-3.77)^2}=6.66\ T$

Magnetic force is given by :

$F=qvB$

Also, F = ma

$a=\dfrac{qvB}{m}$

$a=\dfrac{1.6\times 10^{-19}\times 8\times 10^6\times 6.66}{9.1\times 10^{-31}}$

$a=9.36\times 10^{18}\ m/s^2$

So, the acceleration of the electron is $9.36\times 10^{18}\ m/s^2$. Hence, this is the required solution.