We can use the equation for centripetal force to solve for the magnetic field strength:
F = qvB = (mv^2)/r
where q is the charge of the electron, v is its velocity, B is the strength of the magnetic field, m is the mass of the electron, and r is the radius of the circle.
First, we need to find the velocity of the electron. We can use the relationship between the potential difference (V), energy (E), and charge (q):
E = qV
The energy of the electron is given by E = (1/2)mv^2, so we can set these two equations equal to each other and solve for v:
(1/2)mv^2 = qV
v = sqrt(2qV/m)
Plugging in the given values, we get:
v = sqrt(2 x 1.602 x 10^-19 C x 5.2 x 10^3 V / 9.109 x 10^-31 kg) = 9.89 x 10^6 m/s
Now we can plug in all the values into the first equation and solve for B:
qvB = (mv^2)/r
B = (mv)/qr
B = [(9.109 x 10^-31 kg) x (9.89 x 10^6 m/s)] / [(1.602 x 10^-19 C) x (4.0 x 10^-2 m)]
B = 2.13 x 10^-3 T
Therefore, the magnetic field strength required to keep the electron on a circular path of radius 4.0 cm is 2.13 x 10^-3 T.
An electron is accelerated by s 5.2 kv potential difference how strong a magnetic field must be experienced by the electron if its path is a circle of radius 4.0 cm
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