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