Question

A proton, traveling with a velocity of 4.6 × 106 m/s due east, experiences a magnetic force that has a maximum magnitude of 6.0 × 10-14 N and direction of due south. What are the magnitude and direction of the magnetic field causing the force? If the field is up, then enter a number greater than zero. If the field is down, then enter a number less than zero.

Answer 1

Answer:

$B = 0.0815 T$

since the direction of force is along south so the magnetic field must be upwards

Explanation:

Force on moving charge in magnetic field is given as

$F = q(\vec v \times \vec B)$

here we know this is force on proton

so we have

$q = 1.6 \times 10^{-19} C$

$F = 6.0 \times 10^{-14} (-\hat j) N$

also we know that the velocity of charge is

$v = 4.6 \times 10^6 \hat i m/s$

now from above formula we have

$(6.0 \times 10^{-14}) = (1.6 \times 10^{-19})(4.6 \times 10^6)B$

$B = 0.0815 T$

since the direction of force is along south so the magnetic field must be upwards