Question #310227

Two long parallel wires, 12.0 cm apart are carrying current 1.8 A and 3.5 A respectively in the same directions. The magnetic field at the midpoint between both wires is 5.0 X 10-8 T out of the page. If a proton moves with a velocity of 2.8 X 105 m s-1 at the midpoint between the wires, determine the magnetic force acting on the proton.


1
Expert's answer
2022-03-14T13:07:12-0400

Magnetic force

Bm=μ0i12πdμ0i22πdB_m=\frac{\mu_0i_1}{2\pi d}-\frac{\mu_0i_2}{2\pi d}


Bm=μ0×3.52π×0.06μ0×1.82π×0.06=5.00×108T(approximately)B_m=\frac{\mu_0\times3.5}{2\pi \times0.06}-\frac{\mu_0\times1.8}{2\pi \times0.06}=5.00\times10^{-8}T(approximately)

Magnetic force

F=qvBF=qvB


F=1.6×1019×2.8×105×5×108=2.24×1021NF=1.6\times10^{-19}\times2.8\times10^5\times5\times10^{-8}=2.24\times10^{-21}N


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