Question

*This is a question on a physics document, but beings it's also a mathematics question, it's been tagged as mathematics. Please show your work!*
A 7.12 µC charge is moving at one half the speed of light perpendicular to a magnetic field of 4.02 mT. How large is the force on the charge?


A.) 4.29 N


B.) 32.90 x 10^(1) N


C.) 4.29 x 10^(12) N


D.) 1.00 x 10^(16) N

Answer 1

Answer:

The correct option is A: 4.29 N.

Step-by-step explanation:

The magnetic force is given by:

$F = qv\times B = qvBsin(\theta)$

Where:

q: is the charge = 7.12x10⁻⁶ C

v: is the speed = c/2 = 1.5x10⁸ m/s

B: is the magnetic field = 4.02x10⁻³ T

θ: is the angle = 90° (the speed is perpendicular to the magnetic field)

Hence, the force on the charge is:

$F = qvBsin(90) = 7.12 \cdot 10^{-6} C*1.5 \cdot 10^{8} m/s*4.02 \cdot 10^{-3} T = 4.29 N$

Therefore, the correct option is A: 4.29 N.

I hope it helps you!