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
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.
To find the strength of the magnetic field experienced by the electron, we can use the formula for the centripetal force experienced by a charged particle moving in a magnetic field:
F = qvB,
where F is the centripetal force, q is the charge of the electron, v is its velocity, and B is the magnetic field strength.
The centripetal force can also be related to the acceleration of the electron using Newton's second law:
F = m * a,
where m is the mass of the electron and a is its acceleration.
Since the path of the electron is a circle, the acceleration can be related to its velocity, radius, and centripetal force:
a = v^2 / r.
Combining these equations, we have:
qvB = m * v^2 / r.
Now, let's substitute the given values:
Potential difference (V) = 5.2 kV = 5.2 * 10^3 V,
Radius (r) = 4.0 cm = 4.0 * 10^-2 m.
We know that the potential difference is related to the velocity of the electron using the equation:
V = q * ΔV,
where q is the charge of the electron and ΔV is the potential difference. Rearranging this equation, we have:
v = ΔV / q.
Using the charge of the electron q = 1.6 * 10^-19 C, we can substitute the values and solve for v:
v = (5.2 * 10^3 V) / (1.6 * 10^-19 C) ≈ 3.25 * 10^22 m/s.
Now we can substitute all the known values into the equation above and solve for B:
q * v * B = m * v^2 / r.
Substituting the mass of the electron m = 9.1 * 10^-31 kg, we have:
(1.6 * 10^-19 C) * (3.25 * 10^22 m/s) * B = (9.1 * 10^-31 kg) * (3.25 * 10^22 m/s)^2 / (4.0 * 10^-2 m).
Simplifying and solving for B:
B ≈ [(9.1 * 10^-31 kg) * (3.25 * 10^22 m/s)^2 / (4.0 * 10^-2 m)] / [(1.6 * 10^-19 C) * (3.25 * 10^22 m/s)]
≈ (9.1 * 10^-31 kg * 3.25^2 * 10^44 m^2/s^2) / (1.6 * 10^-19 C * 3.25 * 10^22 m/s)
≈ 47 * 10^-8 Tesla.
Therefore, the magnetic field strength experienced by the electron must be approximately 47 * 10^-8 Tesla.