Question

Express the total magnetic field at point B created by current I1 and I2, BB, in terms of I1, I2, A, and D. Use out of the page as the positive direction.

Answer 1

The total magnetic field
`(\(B_B\))` at point B created by currents
`\(I_1\) and \(I_2\)` can be expressed as follows:

`\[ B_B = (\mu_0)/(4\pi) \left( (I_1)/(D_1) + (I_2)/(D_2) \right) \]`

Step-by-step explanation:

The magnetic field at a point due to a current-carrying wire is given by Ampere's Law. For two parallel straight conductors carrying currents
`\(I_1\) and \(I_2\),` separated by distances
`\(D_1\)` and
`\(D_2\),` the total magnetic field at a point between them (point B) is the sum of the magnetic fields produced by each current.

Ampere's Law states that the magnetic field
`(\(B\))` around a closed loop is proportional to the total current passing through the loop. For a single straight conductor, the formula is
`\(B = (\mu_0)/(4\pi) (I)/(D)\),` where
`\(\mu_0\)` is the permeability of free space,
`\(I\)` is the current, and
`\(D\)` is the distance from the wire.

In the case of two parallel conductors at different distances
`(\(D_1\)` and
`\(D_2\))`from point B, the total magnetic field (\(B_B\)) is the sum of the individual magnetic fields:

`\[ B_B = (\mu_0)/(4\pi) \left( (I_1)/(D_1) + (I_2)/(D_2) \right) \]`

This expression accounts for the contributions of both currents
`\(I_1\)` and
`\(I_2\)` to the magnetic field at point B.