Question

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

Answer 1

Answer:

Magnetic field will be ZERO at the given position

Explanation:

As we know that the magnetic field due to moving charge is given as

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

so here we know that for the direction of magnetic field we will use

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

so we have

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

so magnetic field must be ZERO

So whenever charge is moving along the same direction where the position vector is given then magnetic field will be zero