To solve this problem using Ampere's Law, we first need to understand the concept of Ampere's Law. It states that the line integral of the magnetic field B around a closed loop is equal to the product of the current enclosed by the loop and the permeability of free space (μ0). In other words, it relates the magnetic field around a closed loop to the current passing through the loop.
In this problem, we have two long parallel cylinders with a steady current flowing through them. We want to find the magnetic field at a point inside the inner cylinder (radius a). To do this, we will draw a closed loop around the inner cylinder and apply Ampere's Law.
Step 1: Choosing a closed loop
To apply Ampere's Law, we need to choose a closed loop that encloses the current passing through it. In this case, we will choose a circular loop with a radius r, where r < a. This will ensure that the loop encloses the current passing through the inner cylinder.
Step 2: Determining the magnetic field
Next, we need to determine the magnetic field at a point inside the inner cylinder. According to the Biot-Savart Law, the magnetic field at a point P inside a current-carrying wire is given by:
B = μ0I/2πr
where μ0 is the permeability of free space, I is the current passing through the wire, and r is the distance from the wire to the point P.
In our problem, the current passing through the inner cylinder is I, and the distance from the wire to the point P is r. So, the magnetic field at point P inside the inner cylinder is given by:
B = μ0I/2πr
Step 3: Applying Ampere's Law
Now, we can apply Ampere's Law to the closed loop we chose in Step 1. The line integral of the magnetic field B around the closed loop is given by:
∫B·dl = μ0Ienclosed
where μ0 is the permeability of free space, Ienclosed is the current passing through the loop, and ∫dl is the line integral around the loop.
Since the magnetic field B is constant along the loop, we can take it outside the integral. Also, since the loop encloses the current I, Ienclosed = I. So, the equation becomes:
B∫dl = μ0