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

`B = 0.0815 T`

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

Step-by-step explanation:

Force on moving charge in magnetic field is given as

`F = q(\vec v * \vec B)`

here we know this is force on proton

so we have

`q = 1.6 * 10^(-19) C`

`F = 6.0 * 10^(-14) (-\hat j) N`

also we know that the velocity of charge is

`v = 4.6 * 10^6 \hat i m/s`

now from above formula we have

`(6.0 * 10^(-14)) = (1.6 * 10^(-19))(4.6 * 10^6)B`

`B = 0.0815 T`

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