To solve for the magnetic force exerted on a charge moving perpendicular to a magnetic field, we use the formula for the magnetic force on a moving charge:
[tex]\[ F = q \cdot v \cdot B \][/tex]
where:
- \( F \) is the magnetic force,
- \( q \) is the charge,
- \( v \) is the velocity,
- \( B \) is the magnetic field.
Given:
- The charge \( q = 4.88 \times 10^{-6} \) C (Coulombs),
- The velocity \( v = 265 \) m/s (meters per second),
- The magnetic field \( B = 0.0579 \) T (Teslas).
First, we calculate the magnitude of the magnetic force:
[tex]\[ F = 4.88 \times 10^{-6} \, \text{C} \times 265 \, \text{m/s} \times 0.0579 \, \text{T} \][/tex]
Perform the multiplication step-by-step:
1. Multiply the charge by the velocity:
[tex]\[ 4.88 \times 10^{-6} \, \text{C} \times 265 \, \text{m/s} = 1.2932 \times 10^{-3} \, \text{C} \cdot \text{m/s} \][/tex]
2. Multiply the result by the magnetic field:
[tex]\[ 1.2932 \times 10^{-3} \, \text{C} \cdot \text{m/s} \times 0.0579 \, \text{T} \][/tex]
[tex]\[ = 7.487627999999999 \times 10^{-5} \, \text{N} \][/tex]
Now, we express the result \( 7.487627999999999 \times 10^{-5} \) N in the form \( A \times 10^B \) N:
Here,
- \( A \) is the coefficient,
- \( B \) is the exponent.
So, the magnetic force on the charge can be expressed as:
[tex]\[ 7.487627999999999 \times 10^{-5} \, \text{N} \][/tex]
Therefore, the magnetic force on the charge is:
[tex]\[ \boxed{7.487627999999999 \times 10^{-5} \, \text{N}} \][/tex]
Highlighting the coefficient and the exponent:
[tex]\[ \text{Coefficient: } 7.487627999999999 \][/tex]
[tex]\[ \text{Exponent: } -5 \][/tex]
So, the final answer is:
[tex]\[
7.487627999999999 \times 10^{-5} \, \text{N}
\][/tex]
Here, the coefficient is [tex]\( \boxed{7.487627999999999} \)[/tex] and the exponent is [tex]\( \boxed{-5} \)[/tex].
