A proton moves perpendicularly to a uniform magnetic field B at 3.0 x 10\' m/s a

Question

A proton moves perpendicularly to a uniform magnetic field B at 3.0 x 10' m/s and exhibits an acceleration of 5.0 x 1013 m/s2 in the +x-direction when its velocity is in the +z-direction. Determine the magnitude and direction of the field. T -y direction 10. + -/10 points SerCP7 19.P.017. My Note A wire with a mass of 1.03 g/cm is placed on a horizontal surface with a coefficient of friction of 0.200. The wire carries a current of 1.42 A eastward and moves horizontally to the north. What are the magnitude and the direction of the smallest magnetic field that enables the wire to move in this fashion? ? ---Select--- northward southward Submit Answe westward Practice Another Version upward downward eastward

Explanation / Answer

Force = 1.67*10^-27 kg * 5*10^13 m/s^2 = 8.35*10^-14 N

So, magnitude of field B = 8.35*10^-14 / 1.6*10^-19*3*10^7 = 0.0174 T

For 1m of the wire, to move it, F=I*L*B=0.2*m*g

So, 1.42*1*B = 0.2*0.103*9.81

So, B= 0.14 T and its direction should be downward.

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