A proton moves to the left with a speed of 7.61. 10^degree m/s into a region of

Question

A proton moves to the left with a speed of 7.61. 10^degree m/s into a region of constant magnetic field as shown in the figure below. The magnitude of the magnetic field is 3.95. 10^-2 T and the direction of the field is out of the page. When the proton initially enters the magnetic field, the magnitude of the force on the proton (in N) is 4.810 times 10^-14 You are correct. Computer's answer now shown above. When the proton initially enters the magnetic field, the direction of the force on the proton is

Explanation / Answer

here,

speed , v = 7.61 * 10^6 m/s

B = 0.0395 T

when the proton eneters in the magnetic feild , F = q *(v X B)

F = 1.6 * 10^-19 * ( 7.61 * 10^6 * 0.0395)

F = 4.8 * 10^-14 N

the force on the pparticle is 4.8 * 10^-14 N

and

using right hand rule

the force on the protn is upwards

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