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A proton moves perpendicular to a uniform magnetic field B at a speed of 1.90 x...

Question:

A proton moves perpendicular to a uniform magnetic field B at a speed of 1.90 x 107 m/s and experiences an acceleration of 1.

A proton moves perpendicular to a uniform magnetic field B at a speed of 1.90 x 107 m/s and experiences an acceleration of 1.50 x 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. 1.11e-2 magnitude Apply the expression for the force on a charge moving through a magnetic field to find the field from the acceleration it produces. T direction - Need Help? Read It Master It Submit Anowor Prostico Another Vorsion

Answers

The magnetic field and direction

The magnetic force is f = then 9VB Accelration a=fo = qur m B3 am av B = 1.50x10 x 1.67X10 27 6 xio 19 х 150xjo 7 B = 0.824 X

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