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

Explanation:

Step1: Recall the magnetic force formula

The magnetic force on a moving charge is given by \( F = qvB\sin\theta \), where \( q \) 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.

Step2: Determine the angle \( \theta \)

The charge moves to the right (velocity direction) and the magnetic field is directed into the screen. These two directions are perpendicular, so \( \theta = 90^\circ \) and \( \sin\theta=\sin90^\circ = 1 \).

Step3: Substitute the values into the formula

Given \( q = 5.0\times 10^{-7}\, \text{C} \), \( v = 2.6\times 10^{5}\, \text{m/s} \), \( B = 1.8\times 10^{-2}\, \text{T} \), and \( \sin\theta = 1 \).
\[

$$\begin{align*} F&=qvB\sin\theta\\ &=(5.0\times 10^{-7}\, \text{C})\times(2.6\times 10^{5}\, \text{m/s})\times(1.8\times 10^{-2}\, \text{T})\times1\\ &=(5.0\times2.6\times1.8)\times(10^{-7}\times10^{5}\times10^{-2})\\ &= 23.4\times10^{-4}\\ &= 2.34\times 10^{-3}\, \text{N}\approx2.3\times 10^{-3}\, \text{N} \end{align*}$$

\]

Answer:

\( 2.3\times 10^{-3}\, \text{N} \) (corresponding to the option "2.3 × 10⁻³ N")