Ampere's Law states that the integral of the magnetic field (B) around any closed loop is equal to the product of the permeability of free space (μ0) and the current (I) passing through the loop. Mathematically, it can be expressed as:

∮ B . dl = μ0 * I

Now, let's derive the formula for the magnetic field inside a current-carrying solenoid using Ampere's Law.

Step 1: Consider a solenoid carrying a current I. Let's take a rectangular path abcd inside the solenoid for applying Ampere's law.

Step 2: According to Ampere's law, ∮ B . dl = μ0 * I. Here, I is the total current enclosed by the path abcd.

Step 3: The magnetic field inside the solenoid is uniform and parallel to the axis of the solenoid. So, the line integral of B along the path ab (inside the solenoid) is B * l, where l is the length of the solenoid.

Step 4: The magnetic field outside the solenoid is practically zero. So, the line integral of B along the path bc and da (outside the solenoid) is zero.

Step 5: The line integral of B along the path cd is also zero because the path is perpendicular to the magnetic field.

Step 6: Therefore, the total line integral of B around the path abcd is B * l.

Step 7: According to Ampere's law, B * l = μ0 * I. But here, I is the total current enclosed by the path abcd. If the solenoid has n turns per unit length, then the total current is n * I.

Step 8: Therefore, B * l = μ0 * n * I. From this, we can derive the formula for the magnetic field inside the solenoid as B = μ0 * n * I.

So, the magnetic field inside a current-carrying solenoid is given by B = μ0 * n * I, where n is the number of turns per unit length of the solenoid, I is the current in the solenoid, and μ0 is the permeability of free space.

Question

Ampere's Law states that the integral of the magnetic field (B) around any closed loop is equal to the product of the permeability of free space (μ0) and the current (I) passing through the loop. Mathematically, it can be expressed as:

∮ B . dl = μ0 * I

Now, let's derive the formula for the magnetic field inside a current-carrying solenoid using Ampere's Law.

Step 1: Consider a solenoid carrying a current I. Let's take a rectangular path abcd inside the solenoid for applying Ampere's law.

Step 2: According to Ampere's law, ∮ B . dl = μ0 * I. Here, I is the total current enclosed by the path abcd.

Step 3: The magnetic field inside the solenoid is uniform and parallel to the axis of the solenoid. So, the line integral of B along the path ab (inside the solenoid) is B * l, where l is the length of the solenoid.

Step 4: The magnetic field outside the solenoid is practically zero. So, the line integral of B along the path bc and da (outside the solenoid) is zero.

Step 5: The line integral of B along the path cd is also zero because the path is perpendicular to the magnetic field.

Step 6: Therefore, the total line integral of B around the path abcd is B * l.

Step 7: According to Ampere's law, B * l = μ0 * I. But here, I is the total current enclosed by the path abcd. If the solenoid has n turns per unit length, then the total current is n * I.

Step 8: Therefore, B * l = μ0 * n * I. From this, we can derive the formula for the magnetic field inside the solenoid as B = μ0 * n * I.

So, the magnetic field inside a current-carrying solenoid is given by B = μ0 * n * I, where n is the number of turns per unit length of the solenoid, I is the current in the solenoid, and μ0 is the permeability of free space.

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