The current, I , is +10A, the charge density, h (lambda) , is -1nC/m (=10 -9 C/m

Question

The current,I, is +10A, the charge density,h (lambda), is -1nC/m (=10-9C/m) (note that it is negative), and the distance between the wires is 40cm. At the instant shown, a proton with chargeq= 1.6 Ý 10-19C is moving into the page with a speedv= 106m/s. Ignoring gravity, what is the magnitude and direction of the net force it feels at that time?

The current,I, is +10A, the charge density, h (lambda), is -1nC/m (=10 -9C/m) (note that it is negative), and the distance between the wires is 40cm. At the instant shown, a proton with chargeq= 1.6 10 -19C is moving into the page with a speedv= 10 6m/s. Ignoring gravity, what is the magnitude and direction of the net force it feels at that time?

Explanation / Answer

B at given point considering the point to be at the middle of the wires = muo x I / (2 x pi x d/2)

= 4 x pi x 10^-7 x 10 / (2 x pi x 0.2m)

= 1 x 10^-5 T.

E at the same point = 2x k x h / (d/2)

= 2 x 8.98 x 10^9 x 1 x 10^-9 / 0.2

= 89.8 N/C

Fo net force acting on the proton = Bqv + Eq

= [1 x 10^-5 x 1.6 x 10^-19 x 10^6] + [89.9 x 1.6 x 10^-19]

= 10 x 10^-19 + 143.84 x 10^-19

= 159.84 x 10^-19 N.

Angle with respect to Electric field direction, i.e. upward = tan^(-1) [10x 10^-19 / 143.84 x 10^-19]

= 3.97 degree with Electric field direction that is upward into the page.

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