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…

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asked May 28, 2020 163k views

1 Answer

Yahyazini · Answer 1 · 2020-06-03T04:58:14+0000

Answer:

Acceleration,
$a=9.36* 10^(18)\ m/s^2$

Step-by-step explanation:

It is given that,

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

Charge on an electron,
$q=1.6* 10^(-19)\ C$

Mass of electron,
$m=9.1* 10^(-31)\ kg$

Magnetic field,
B=5.5i-3.7j

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

Magnetic force is given by :

F=qvB

Also, F = ma

a=(qvB)/(m)

a=(1.6* 10^(-19)* 8* 10^6* 6.66)/(9.1* 10^(-31))

$a=9.36* 10^(18)\ m/s^2$

So, the acceleration of the electron is
$9.36* 10^(18)\ m/s^2$ . Hence, this is the required solution.