Final answer:
The magnitude of Earth's magnetic field can be calculated using the formula B = mv/(qr), where m is the mass of the electron, v is the velocity of the electron, q is the charge of the electron, and r is the radius of the circular path.
Step-by-step explanation:
The question involves calculating the magnitude of Earth's magnetic field based on the known speed of an electron and the radius of its circular path when moving perpendicular to the magnetic field. This can be answered using the formula that relates the magnetic force acting on a moving charge to the magnetic field: F = qvB. Here, F is the magnetic force equal to the centripetal force (Fc), q is the charge of the electron, v is the speed of the electron, and B is the magnetic field strength. Since Fc = mv²/r, where m is the mass of the electron and r is the radius of the circular path, we can rearrange the equation and solve for B: B = mv/(qr).
Given the electron's speed (v = 5.84×10⁵ m/s) and the circular path radius (r = 6.79×10⁻² m), along with the electron's mass (m = 9.11×10⁻³¹ kg) and charge (q = -1.60×10⁻¹⁹ C), we can calculate the magnetic field strength. Plugging in these values, the magnitude of Earth's magnetic field that would cause this trajectory is determined. It's important to use the absolute value of the electron's charge since we're interested in the magnitude of the magnetic field.