Suppose that a nascar race car is moving to the right with a constant velocity of +82m/s. what is the average acceleration of th
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Answer:
3A is the larger of the two currents.
Explanation:
Let the currents in the two wires be I₁ and I₂
given:
Magnitude of the electric field, B = 4.0μT = 4.0×10⁻⁶T
Distance, R = 10cm = 0.1m
Ratio of the current = I₁ : I₂ = 3 : 1
Now, the magnitude of a magnetic field at a distance 'R' due to the current 'I' is given as
Where is the magnitude constant = 4π×10⁻⁷ H/m
Thus, the magnitude of a magnetic field due to I₁ will be
given,
B = B₁ - B₂ (since both the currents are in the same direction and parallel)
substituting the values of B, B₁ and B₂
we get
4.0×10⁻⁶T = -
or
4.0×10⁻⁶T =
also
⇒
substituting the values in the above equation we get
4.0×10⁻⁶T =
⇒
also
⇒
⇒
Hence, the larger of the two currents is 3A