Calculate the magnetic force on an electron in a magnetic field of flux density 10T, with a velocity of 3 x 10…

1. Analyze the Question

We are asked to calculate the magnetic force ($F$) acting on a moving electron within a magnetic field. We are given:

• Magnetic flux density ($B$) = $10 \, \text{T}$
• Velocity of the electron ($v$) = $3 \times 10^7 \, \text{m/s}$
• Angle ($\theta$) = $60^\circ$
• Charge of an electron ($e$) = $1.6 \times 10^{-19} \, \text{C}$

2. Explain the Concept

The magnetic force on a moving charge is given by the Lorentz force formula:

Where:

• $F$ is the magnetic force in Newtons (N).
• $q$ is the charge of the particle (in this case, the electron $e$).
• $v$ is the velocity of the particle.
• $B$ is the magnetic flux density.
• $\theta$ is the angle between the velocity vector and the magnetic field lines.

3. Show the Solution

Substitute the given values into the formula:

• $F = (1.6 \times 10^{-19}) \times (3 \times 10^7) \times (10) \times \sin(60^\circ)$

First, calculate the product of the constants:

• $1.6 \times 3 \times 10 = 48$
• $10^{-19} \times 10^7 \times 10^1 = 10^{-11}$
• So, $qvB = 48 \times 10^{-11}$

Next, include $\sin(60^\circ)$:

• $\sin(60^\circ) \approx 0.866$
• $F = 48 \times 10^{-11} \times 0.866$
• $F = 41.568 \times 10^{-11} \, \text{N}$

To express this in standard scientific notation:

• $F \approx 4.1568 \times 10^{-11} \, \text{N}$

Rounding to two decimal places, we get $4.16 \times 10^{-11} \, \text{N}$.

4. Identify the Answer

Comparing our result to the given options, it matches option C.

5. Tip

When solving magnetism problems, always check the angle. If the particle moves parallel to the field ($\theta = 0^\circ$), the force is zero because $\sin(0) = 0$. If it moves perpendicular ($\theta = 90^\circ$), the force is at its maximum because $\sin(90) = 1$. Remembering these extremes helps you verify if your answer is reasonable!