Question

A point charge of 5.0 × 10–7 C moves to the right at 2.6 × 105 m/s in a magnetic field that is directed into the screen and has a field strength of 1.8 × 10–2 T.

What is the magnitude of the magnetic force acting on the charge?

0 N
2.3 × 10–3 N
23 N
2.3 × 1011 N

Answer 1

To find the magnitude of the magnetic force acting on a moving charge in a magnetic field, we can use the formula:

\[
F = qvB \sin(\theta)
\]

Where:
- \( F \) is the magnetic force,
- \( q \) is the charge,
- \( v \) is the velocity of the charge,
- \( B \) is the magnetic field strength,
- \( \theta \) is the angle between the velocity of the charge and the magnetic field.

In this case:
- The charge \( q = 5.0 \times 10^{-7} \, \text{C} \)
- The velocity \( v = 2.6 \times 10^{5} \, \text{m/s} \)
- The magnetic field \( B = 1.8 \times 10^{-2} \, \text{T} \)
- Since the charge is moving to the right and the magnetic field is directed into the screen, the angle \( \theta = 90^\circ \), hence \( \sin(90^\circ) = 1 \).

Plugging the values into the formula, we get:

\[
F = (5.0 \times 10^{-7} \, \text{C})(2.6 \times 10^{5} \, \text{m/s})(1.8 \times 10^{-2} \, \text{T}) \cdot 1
\]

Calculating this step-by-step:

1. First, calculate \( qv \):

\[
5.0 \times 10^{-7} \times 2.6 \times 10^{5} = 1.3 \times 10^{-1} \, \text{C m/s}
\]

2. Now, multiply by \( B \):

\[
F = (1.3 \times 10^{-1}) \times (1.8 \times 10^{-2}) = 2.34 \times 10^{-3} \, \text{N}
\]

Thus, rounding this value gives approximately:

\[
F \approx 2.3 \times 10^{-3} \, \text{N}
\]

So the magnitude of the magnetic force acting on the charge is:

\[
\boxed{2.3 \times 10^{-3} \, \text{N}}
\]