A proton moves perpendicular to a uniform magnetic field at a speed of 2.00 107

Question

A proton moves perpendicular to a uniform magnetic field at a speed of 2.00 107 m/s and experiences an acceleration of 2.90 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.
magnitude =__________T
Direction = +x, -X, +y, -y, +z, or -Z???????


A wire having mass per unit length of 0.480 g/cm carries a 2.60 A current horizontally to the south. What are the direction and magnitude of the minimum magnetic field needed to lift this wire vertically upward?
=________T
in the southward, northward, eastward, or westward direction????

Explanation / Answer

F = qVB = 1.6*10^19 * 2*10^7 * B = ma 1.6*10^-19 * 2*10^7 * B = 1.67*10^-27 * 2.90*10^13 B = 0.015134 T F = IBL = mg B = mg / (IL) B = 0.48*10^-3 *9.8 / (10^-2 *2.6) B = 0.1809T westwards

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