Answer:
3.3 mT
Explanation:
First of all, we need to find the strength of the electric field between the two parallel plates.
We have:
[tex]\Delta V=200 V[/tex] (potential difference between the two plates)
[tex]d=1.0 cm=0.01 m[/tex] (distance between the plates)
So, the electric field is given by
[tex]E=\frac{\Delta V}{d}=\frac{200 V}{0.01 m}=2\cdot 10^4 V/m[/tex]
Now we want the electron to pass between the plates without being deflected; this means that the electric force and the magnetic force on the electron must be equal:
[tex]F_E = F_B\\qE=qvB[/tex]
where
q is the electron charge
E is the electric field strength
v is the electron's speed
B is the magnetic field strength
In this case, we know the speed of the electron: [tex]v=6.0\cdot 10^6 m/s[/tex], so we can solve the formula to find B, the magnetic field strength:
[tex]B=\frac{E}{v}=\frac{2\cdot 10^4 V/m}{6.0\cdot 10^6 m/s}=0.0033 T=3.3 mT[/tex]
The magnetic field strength will allow the electron to pass between the plates without being deflected is 0.0033 T.
The electric field strength of the electron is calculated as follows;
E = V/d
E = 200/(0.01)
E = 20,000 V/m
The magnetic field strength is related to electric field in the following formula;
qvB = qE
vB = E
B = E/v
B = (20,000)/(6 x 10⁶)
B = 0.0033 T
Thus, the magnetic field strength will allow the electron to pass between the plates without being deflected is 0.0033 T.
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