A \(10\ eV\) electron is circulating in a plane at right angles to a uniform field at magnetic induction \({10}^{- 4}\ Wb/m^2\) (= \(1.0\) gauss), the orbital radius of electron is

Question

A long straight current-carrying wire is placed in a uniform magnetic field of strength \( B = 0.5 \, \text{T} \). If the current in the wire is \( I = 2 \, \text{A} \) and the wire makes an angle of \( 30^\circ \) with the magnetic field, find the force per unit length on the wire.

Answer 1

To find the orbital radius of an electron circulating in a magnetic field, we use the formula for the radius of the path of a charged particle moving perpendicular to a uniform magnetic field:

r = \frac{mv}{qB}

Where:

r is the radius of the orbit.
m is the mass of the electron (9.11 x 10^-31 kg).
v is the velocity of the electron.
q is the charge of the electron (1.6 x 10^-19 C).
B is the magnetic field (1.0 gauss = 10^-4 Wb/m²).

First, we need to calculate the velocity v of the electron using its kinetic energy. The kinetic energy of the electron (K.E.) is given by:

K.E. = \frac{1}{2}mv^2 = 10\ eV

Convert 10 eV into joules:

1\ eV = 1.6 \times 10^{-19}\ J.

Thus, 10\ eV = 1.6 \times 10^{-18}\ J.

We set up the equation:

\frac{1}{2}mv^2 = 1.6 \times 10^{-18}

Solving for v:

v^2 = \frac{2 \times 1.6 \times 10^{-18}}{9.11 \times 10^{-31}}

v = \sqrt{\frac{3.2 \times 10^{-18}}{9.11 \times 10^{-31}}}

v ≈ 1.87 \times 10^6 m/s

Now, substitute these values into the radius formula:

r = \frac{(9.11 \times 10^{-31})(1.87 \times 10^6)}{1.6 \times 10^{-19} \times 10^{-4}}

r = \frac{1.704 \times 10^{-24}}{1.6 \times 10^{-23}}

r ≈ 0.107\ m = 10.7\ cm

Rounding to two decimal places, the radius is approximately 11 cm, which matches the given correct answer.

Answer 2