Question

What is the velocity of a beam of electrons that goes undeflected when passing through perpendicular electric and magnetic fields of magnitude 6800 V/m and 6.9×10−3 T , respectively?

Answer 1

Answer:

Velocity of electron will be equal to $9.855\times 10^5m/sec$

Explanation:

We have given electric field is equal to E = 6800 volt/m

Magnetic field is equal to $B=6.9\times 10^{-3}T$

We have to find the velocity of the electron

We know that velocity of electron is equal to ratio of electric field and magnetic field

So $v=\frac{E}{B}=\frac{6800}{6.9\times 10^{-3}}=9.855\times 10^5m/sec$

Velocity of electron will be equal to $9.855\times 10^5m/sec$

Answer 2

Answer:

The velocity of a beam of electrons is $9.85\times 10^{-5}\ m/s$.

Explanation:

Given that,

Electric field, $E=6800\ V/m$

Magnetic field,    $B=6.9\times 10^{-3}\ T$

When electric and magnetic field both act on the particle, the net force is given by :

$F=(qE+qv\times B)$

If the electron moves undeflected, F = 0

So,

$E=v\times B\\\\v=\dfrac{E}{B}\\\\v=\dfrac{6800}{6.9\times 10^{-3}}\\\\v=9.85\times 10^5\ m/s$

So, the velocity of a beam of electrons is $9.85\times 10^{-5}\ m/s$. Hence, this is the required solution.