Question

a proton moves perpendicular to a uniform magnetic field b at a speed of 1.25 107 m/s and experiences an acceleration of 2.45 1013 m/s2 in the positive x direction when its velocity is in the positive z direction. determine the magnitude and direction of the field.

Answer 1

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

How we do it?

The magnetic field must first be in the opposite direction from acceleration, since acceleration occurs in the positive x-direction.

We know that the magnitude of the Lorentz force F is; F = q v B sinθ

So, B = F/(qvsinθ)

F = ma.

Speed(v) = 1.25 x10^(7) m/s

Acceleration (a) = 2.45 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 2.45 x10^(13)] / [ {1.602×10^(−19)} x {1.00 x10^(7)} x sin 90]

So ,B = 2.55 x 10^(-2) T.

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

To know more about Lorentz force, visit-https://brainly.com/question/28211140

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