Question

a point charge of 6.8 μc moves at 6.5 × 10⁴ m/s at an angle of 15° to a magnetic field that has a field strength of 1.4 t. what is the magnitude of the magnetic force acting on the charge? ○ 1.6 × 10⁻¹ n ○ 6.0 × 10⁻² n ○ 1.6 × 10⁵ n ○ 6.0 × 10⁶ n

Explanation:

Step1: Recall the formula for magnetic force on a moving charge

The formula for the magnetic force \( F \) on a moving charge \( q \) is \( F = qvB\sin\theta \), where \( v \) is the velocity of the charge, \( B \) is the magnetic field strength, and \( \theta \) is the angle between the velocity and the magnetic field.

Step2: Convert the charge to coulombs

The charge \( q = 6.8\ \mu\text{C} = 6.8\times 10^{- 6}\ \text{C} \).

Step3: Substitute the values into the formula

We have \( q = 6.8\times 10^{-6}\ \text{C} \), \( v=6.5\times 10^{4}\ \text{m/s} \), \( B = 1.4\ \text{T} \), and \( \theta=15^{\circ} \).
First, calculate \( \sin(15^{\circ})\approx\sin(15^{\circ})\approx0.2588 \)
Then, \( F=(6.8\times 10^{-6}\ \text{C})\times(6.5\times 10^{4}\ \text{m/s})\times(1.4\ \text{T})\times\sin(15^{\circ}) \)
Calculate the product step by step:
\( (6.8\times 6.5\times 1.4\times0.2588)\times10^{-6 + 4} \)
\( 6.8\times6.5 = 44.2 \)
\( 44.2\times1.4=61.88 \)
\( 61.88\times0.2588\approx16.0 \)
\( 10^{-6 + 4}=10^{-2} \)
So \( F\approx16.0\times 10^{-2}\ \text{N}=1.6\times 10^{-1}\ \text{N} \)

Answer:

\( 1.6\times 10^{-1}\ \text{N} \) (corresponding to the option \( 1.6\times 10^{-1}\ \text{N} \))