Two long, parallel wires carry unequal currents in the same direction. the ratio of the currents is 3 to 1. the magnitude of the magnetic field at a point in the plane of the wires and 10 cm from each wire is 4.0 μt. what is the larger of the two currents?

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

a.   mass number from atomic number

get the total number of pieces of fruit and divide it by 4 to get the average. the result is the answer to the problem. 14 total number of fruits and there are 4 members. 14 divided by 4 is equals to 3.5.