A proton moves along the x-axis with vx=1.0×107m/s. As it passes the origin, what are the strength and direction of the magnetic

Question

3 months ago

11

field at the (1 cm, 0 cm, 0 cm) position? Give your answer using unit vectors.

1 answer:

inna [3.1K] · Answer 1 · 2026-03-16T05:08:36+03:00

Answer:

At this position, the magnetic field equals ZERO

Explanation:

The magnetic field produced by a moving charge is described as

$B = \frac{\mu_0 qv sin\theta}{4\pi r^2}$

Here, we determine the direction of the magnetic field using

$\hat B = \hat v \times \hat r$

Thus, we find

$\hat B = \hat i \times (\hat i + 0\hat j + 0\hat k)$

Leading to a magnetic field of ZERO

Consequently, when the charge moves in the same line as the given position vector, the magnetic field will be nonexistent