# An electron moves in a magnetic field along a circle with a radius of 10 cm. Magnetic induction is 20 mTL.

An electron moves in a magnetic field along a circle with a radius of 10 cm. Magnetic induction is 20 mTL. Determine the kinetic energy of the electron. The mass of an electron is 9.1 * 10 ^ (- 30) kg. q = -1.6 * 10 ^ (- 19) Cl.

Given:
R = 10 cm;
B = 20 mTL;
m electron = 9.1 * 10 ^ (- 30) kg;
q = -1.6 * 10 ^ (- 19) Cl.
Find: E is kinetic.
Decision:
In order to solve the problem, you will first need to use Newton’s second law. It looks like this: Fr = F Lorentz. We also need a second formula: m * a = q * v * B.
Let us express the acceleration a from it. Then we get a = m * v ^ 2 / r.
Next, we write down the acceleration formula. a = v ^ 2 / r.
We can equate the right-hand sides of the two formulas. We get:
m * v ^ 2 / r = q * v * B.
From the resulting equation, you need to express the speed.
v = q * B * r / m.
Now we can write down the formula E kinetic. It looks like this:
E kinetic = m * v ^ 2/2.
In it, we replace the speed with the right side of the above formula and get: v = m * (q * B * r / m) ^ 2/2 /
Next, we carry out transformations to make it easier to count: E kinetic = (q * B * r) ^ 2 / (2 * m).
Now we substitute the values ​​and find the answer:
E kinetic = 562.6 * 10 ^ -16 J.
Answer: 562.6 * 10 ^ -16 J

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