Question

The magnitude of the magnetic force on a particle with charge \(-2 e\) moving with speed \(v=1.00 \cdot 10^{5} \mathrm{~m} / \mathrm{s}\) is \(3.00 \cdot 10^{-18} \mathrm{~N}\). What is the magnitude of the magnetic field component perpendicular to the direction of motion of the particle?

Short answer

Answer: The magnitude of the magnetic field component perpendicular to the direction of motion of the particle is 9.375 x 10^{-5} T.

Step-by-step solution

Write down the given information

We are given the following information: - Charge of the particle: -2e (where e is the elementary charge, e = 1.6 x 10^{-19} C) - Speed of the particle: v = 1.00 x 10^5 m/s - Magnitude of the magnetic force: F = 3.00 x 10^{-18} N - Angle between magnetic field and velocity: θ = 90 degrees (sin(90) = 1)

Write down the formula for the magnetic force acting on a moving charged particle and rearrange for B

The magnetic force acting on a moving charged particle is given by the equation: F = qvBsinθ Since we know that θ = 90 degrees, and sin(90) = 1, the equation simplifies to: F = qvB We need to find B, so we can rearrange the equation for B: B = F / (qv)

Plug in the given values and solve for B

Now we have everything we need to solve for B. Plug in the given values into the rearranged equation: B = (3.00 x 10^{-18} N) / [(-2)(1.6 x 10^{-19} C)(1.00 x 10^5 m/s)] B = (3.00 x 10^{-18} N) / (-3.2 x 10^{-14} C m/s) B = 9.375 x 10^{-5} T Since we are only interested in the magnitude of the magnetic field component, we will report the absolute value: B = 9.375 x 10^{-5} T (in magnitude)

State the final answer

The magnitude of the magnetic field component perpendicular to the direction of motion of the particle is 9.375 x 10^{-5} T.