Question

a charged particle moves at 2.5 × 10⁴ m/s at an angle of 25° to a magnetic field that has a field strength of 8.1 × 10⁻² t. if the magnetic force is 7.5 × 10⁻² n, what is the magnitude of the charge?
○ 3.7 × 10⁻⁵ c
○ 4.1 × 10⁻⁵ c
○ 8.8 × 10⁻⁵ c
○ 1.0 × 10⁻⁴ c

Explanation:

Step1: Recall the magnetic force formula

The formula for the magnetic force on a charged particle is \( F = qvB\sin\theta \), where \( F \) is the magnetic force, \( q \) is the charge, \( v \) is the velocity, \( B \) is the magnetic field strength, and \( \theta \) is the angle between the velocity and the magnetic field. We need to solve for \( q \), so we rearrange the formula to \( q=\frac{F}{vB\sin\theta} \).

Step2: Identify the given values

We are given:

• \( F = 7.5\times 10^{-2}\, \text{N} \)
• \( v = 2.5\times 10^{4}\, \text{m/s} \)
• \( B = 8.1\times 10^{-2}\, \text{T} \)
• \( \theta = 25^\circ \)

First, calculate \( \sin(25^\circ) \). Using a calculator, \( \sin(25^\circ)\approx 0.4226 \).

Step3: Substitute the values into the formula

Substitute \( F \), \( v \), \( B \), and \( \sin\theta \) into the formula for \( q \):

\[
q=\frac{7.5\times 10^{-2}}{(2.5\times 10^{4})\times(8.1\times 10^{-2})\times 0.4226}
\]

First, calculate the denominator:

\( (2.5\times 10^{4})\times(8.1\times 10^{-2}) = 2.5\times8.1\times 10^{4 - 2}=20.25\times 10^{2}=2025 \)

Then, multiply by \( 0.4226 \): \( 2025\times0.4226\approx 2025\times0.4226 = 855.765 \)

Now, calculate the numerator divided by the denominator:

\( q=\frac{7.5\times 10^{-2}}{855.765}\approx\frac{0.075}{855.765}\approx 8.76\times 10^{-5}\, \text{C}\approx 8.8\times 10^{-5}\, \text{C} \)

Answer:

\( 8.8\times 10^{-5}\, \text{C} \) (corresponding to the option "8.8 × 10⁻⁵ C")