Two long parallel wires 6.50cm apart carry 18.5A currents in the same direction. Determine the magnetic field strength at a point 12.0cm from one wire and 13.6cm from the other.

Explanation:

The magnetic field produced by a current-carrying wire is:

where

is the vacuum permeability

I is the current in the wire

r is the distance from the wire

The direction of the field is given by the right-hand rule: the thum is placed in the same direction of the current, and the other fingers "wrapped" around the thumb gives the direction of the field.

First of all, here we are finding the magnetic field strength at a point 12.0cm from one wire and 13.6cm from the other: since the two wires are 6.50 cm apart, this means that the point at which we are calculating the field is located either on the left or on the right of both wires. So, by using the right-hand rule for both, we see that both fields go into the same direction (because the currents in the two wires have same direction).

Therefore, the net magnetic field will be the sum of the two magnetic fields.

For wire 1, we have:

So the field is

For wire 2, we have:

So the  field is

Therefore, the total magnetic field is

we have been given a quadratic function and we need to restrict the domain such that it becomes a one to one function.

we know that vertex of this quadratic function occurs at (5,2).

further, we know that range of this function is .

if we restrict the domain of this function to either or , it will become one to one function.

let us know find its inverse.

upon interchanging x and y, we get:

let us now solve this function for y.

hence, the inverse function would be if we restrict the domain of original function to and the inverse function would be if we restrict the domain to .