Question

An electron at point A has a speed v₀ of 1.41 x 10⁶ m/s. Find

A. The magnitude and direction of the magnetic field that will cause the lectron to follow the semicircular path from A to B, and;

B. The time required for the electron to move from A to B.

Answer 1

Speed of an electron \(v_{0}=1.41 \times 10^{6} \mathrm{~m} / \mathrm{s}\)

Radius of circular orbit \(R=5.0 \mathrm{~cm}\)

(a) Let \(B\) be the strength of a magnetic field.

The equation for the motion of an electron is \(F=q v B\)

\(=\frac{m v^{2}}{R}\)

\(B=\frac{m v}{q R}\)

where the mass of the electron \(m=9.1 \times 10^{-31} \mathrm{~kg}\)

and charge of the electron \(q=1.6 \times 10^{-19} \mathrm{C}\)

From these values, the strength of the magnetic field

\(B=\frac{\left(9.11 \times 10^{-31} \mathrm{~kg}\right)\left(1.41 \times 10^{6} \mathrm{~m} / \mathrm{s}\right)}{\left(1.60 \times 10^{-19} \mathrm{C}\right)\left(5.0 \times 10^{-2} \mathrm{~m}\right)}\)

\(=1.60 \times 10^{-4} \mathrm{~T}\)

(b) Distance traveled by the electron from point \(A\) to \(B \quad l=\pi R\)

Time required by the electron to move from point \(A\) to B \(t=\frac{l}{v_{0}}\)

\(=\frac{\pi R}{v_{0}}\)

\(=\frac{\pi \times\left(5.0 \times 10^{-2} \mathrm{~m}\right)}{\left(1.41 \times 10^{6} \mathrm{~m} / \mathrm{s}\right)}\)

\(=1.114 \times 10^{-7} \mathrm{~s}\)