48.4k views
10 votes
*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

User Kasur
by
7.3k points

1 Answer

4 votes

Answer:

The correct option is A: 4.29 N.

Explanation:

The magnetic force is given by:


F = qv* 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!

User Pajevic
by
8.2k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories