Question

At a certain location, the horizontal component of the earth's magnetic field is 3.1 x 10⁻⁵ T due north. A proton moves eastward with just the right speed, so the magnetic force on it balances its weight. Find the speed of the proton.

Answer 1

The speed of the proton is approximately 8.60 x 10^5 m/s.

Step-by-step explanation:

To find the speed of the proton, we need to balance the magnetic force with its weight. The magnetic force on the proton is given by the equation Fm = qvB, where q is the charge of the proton, v is its velocity, and B is the magnetic field.

If the magnetic force balances the weight of the proton, then Fm = mg, where m is the mass of the proton and g is the acceleration due to gravity.

Combining these equations, we have qvB = mg. We can rearrange this equation to solve for v:

v = (mg) / (qB)

Substituting the given values, we have v = (1.67 x 10^-27 kg * 9.8 m/s^2) / (1.6 x 10^-19 C * 3.1 x 10^-5 T).

By calculating this equation, we find that the speed of the proton is approximately 8.60 x 10^5 m/s.