Protons move in a circle of radius 7.40 cm in a 0.552-T magnetic field. What val

Question

Protons move in a circle of radius 7.40 cm in a 0.552-T magnetic field. What value of electric field could make their paths straight? In what direction must it point? The direction of the electric field is in the same direction to the velocity. The direction of the electric field is perpendicular to both the velocity and the magnetic field, and in the opposite direction to the magnetic force on the protons. The direction of the electric field is in the opposite direction to the magnetic field. The direction of the electric field is in the opposite direction to the velocity. The direction of the electric field is perpendicular to both the velocity and the magnetic field, and in the same direction to the magnetic force on the protons. The direction of the electric field is in the same direction to the magnetic field.

Explanation / Answer

radius r = 7.4 cm = 7.4 x10 -2 m

Magnetic field B = 0.552 T

Charge of proton q = 1.6 x10 -19 C

Mass of proton m = 1.67 x10 -27 kg

In Magnetic field , Bvq = mv 2 / r

Bq = mv/r

From this v = Bqr / m

= (0.552)(1.6 x10 -19 )( 7.4 x10 -2)/(1.67 x10 -27)

= 3.913x10 6 m/s

We know for straight line path ,

force due to magnetic field = force due to electric field

Bvq = Eq

Bv = E

E = 0.552 x 3.913x10 6

     = 2.16x10 6 N/C

(B). option(2) is correct

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