A proton moves perpendicular to a uniform magnetic field B at a speed of 1.00 x10^7 m/s and experiences an acceleration of 2.00 x10^13 m/s^2. in the positive x-direction when its velocity is in the positive z-direction. Determine the magnitude and direction of the field for which the magnitude of the field is a minimium

So, Magnitude of the field (B) = 2.09 x 10^ (-2) T.

It's in the negative direction since acceleration is in positive direction.

Explanation:

First of all, since acceleration is in positive x - direction, the magnetic field must be in negative y - direction.

We know that The magnitude of the Lorentz force F is; F = qvB sinθ

So, B = F / (qvsinθ)

F = ma.

Speed (v) = 1.00 x10^ (7) m/s

acceleration (a) = 2.00 x10^ (13) m/s^ (2)

Mass of proton = 1.673 * 10^ (-27) kilograms

q (elementary charge of proton) = 1.602*10^ (-19)

Since right hand thumb rule, θ = 90°

So; B = [1.673 * 10^ (-27) x 2.00 x10^ (13) ] / [ {1.602*10^ (-19) } x {1.00 x10^ (7) } x sin 90]

So, B = 2.09 x 10^ (-2) T.

It's in the negative direction since acceleration is in positive direction.