Answered

A charge feels a [tex]2.89 \times 10^{-7} N[/tex] force when it moves at [tex]288 \, m/s[/tex] perpendicular [tex]\left(90^{\circ}\right)[/tex] to a magnetic field of [tex]2.77 \times 10^{-5} T[/tex]. How big is the charge?

[tex][?] \times 10^{[?]} C[/tex]



Answer :

To find the charge, we can use the formula for the magnetic force acting on a moving charge:

[tex]\[ F = q \cdot v \cdot B \cdot \sin(\theta) \][/tex]

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 and the magnetic field.

Given the problem:
- \( F = 2.89 \times 10^{-7} \) N
- \( v = 288 \) m/s
- \( B = 2.77 \times 10^{-5} \) T
- \( \theta = 90^{\circ} \)

Since the angle \(\theta\) is \(90^{\circ}\), \(\sin(90^{\circ}) = 1\). Thus, the formula simplifies to:

[tex]\[ F = q \cdot v \cdot B \][/tex]

Rearrange the formula to solve for \( q \):

[tex]\[ q = \frac{F}{v \cdot B} \][/tex]

Substitute the given values into the formula:

[tex]\[ q = \frac{2.89 \times 10^{-7} \, \text{N}}{288 \, \text{m/s} \times 2.77 \times 10^{-5} \, \text{T}} \][/tex]

[tex]\[ q = \frac{2.89 \times 10^{-7}}{7.9776 \times 10^{-3}} \][/tex]

[tex]\[ q \approx 3.6226434015242685 \times 10^{-5} \, \text{C} \][/tex]

Thus, the charge \( q \) is approximately:
[tex]\[ 3.62 \times 10^{-5} \, \text{C} \][/tex]

So, the charge that experiences the given force under the specified conditions is:
[tex]\[ 3.62 \times 10^{-5} \, \text{C} \][/tex]