contestada

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

Respuesta :

Answer:

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.

Otras preguntas