Electron in Magnetic Field Problem - Detailed Solution

Electron in Magnetic Field Problem

Question: A long straight wire carries a current and produces a magnetic field of \( 2 \times 10^{-4}\,\text{T} \) at a perpendicular distance of 5 cm from the wire. An electron situated at 5 cm from the wire moves with a velocity of \( 10^7\,\text{m/s} \) towards the wire, moving perpendicularly to it. (Charge on the electron is \( 1.6 \times 10^{-19}\,\text{C} \)). Find the force experienced by the electron.

(a) \( 3.2\,\text{N} \)

(b) \( 3.2 \times 10^{-16}\,\text{N} \)

(c) \( 1.6 \times 10^{-16}\,\text{N} \)

(d) Zero

Detailed Step-by-Step Explanation

Step 1: Identify the Given Values
The magnetic field at a distance of 5 cm is given as: \[ B = 2 \times 10^{-4}\,\text{T} \] The electron's velocity is: \[ v = 10^7\,\text{m/s} \] The charge on the electron is: \[ q = 1.6 \times 10^{-19}\,\text{C} \]

Step 2: Apply the Lorentz Force Law
The magnetic force \( F \) on a moving charge is given by: \[ F = q v B \sin \theta \] Since the electron is moving perpendicularly to the magnetic field (i.e., \(\theta = 90^\circ\)), \(\sin 90^\circ = 1\). Thus: \[ F = q v B \]

Step 3: Substitute the Values
Substitute \( q = 1.6 \times 10^{-19}\,\text{C} \), \( v = 10^7\,\text{m/s} \), and \( B = 2 \times 10^{-4}\,\text{T} \) into the equation:

\[ F = (1.6 \times 10^{-19}) \times (10^7) \times (2 \times 10^{-4}) \]

Step 4: Simplify the Calculation
Calculate the product of the constants: \[ 10^7 \times 2 \times 10^{-4} = 2 \times 10^3 \] Then: \[ F = 1.6 \times 10^{-19} \times 2 \times 10^3 = 3.2 \times 10^{-16}\,\text{N} \]

Final Answer: The force experienced by the electron is \( 3.2 \times 10^{-16}\,\text{N} \), which corresponds to option (b).