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a proton is at the position (2, 1) with a speed of 750,000 m/s and is moving in the positive x-direction. what is the magnitude of the magnetic field created by this proton at the position (5, 5)? all positions are given in si units.

User HYS
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The magnitude of the magnetic field created by the proton at the given position can be calculated using the formula B = μ0qv/2πr. The calculated magnitude of the magnetic field is approximately 1.92 x 10^{-6} T.

The magnitude of the magnetic field created by the proton can be calculated using the formula:


B = μ0qv/2πr

Where:


B is the magnitude of the magnetic field

μ0 is the permeability of free space (4π×10-7 Tm/A)

q is the charge of the proton (1.602 × 10-19 C)

v is the speed of the proton (750,000 m/s)

r is the distance between the proton and the point where the magnetic field is being measured (distance between (2,1) and (5,5))


Using the given values and calculating:

B = (4 × 3.14 × 10-7 × 1.602 × 10-19 C × 750,000 m/s) / (2 × 3.14 × distance)

Where distance = √((5-2)2 + (5-1)2)

distance = √((3)2 + (4)2) = √ (9 + 16) = √ 25 = 5m


Substituting back into the equation:


B = (4 × 3.14 × 10-7 × 1.602 × 10-19 C × 750,000 m/s) / (2 × 3.14 × 5)


Finally, solving the equation:

B ≈ 1.92 x 10-6 T

User Mrgoos
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