Two large conducting planes carrying current perpendicular to x axis and placed at b, 0 and 20, 0 as shown in figure current per unit with in both the plane is cm and current is flowing in outward direction variation of magnetic induction as function of x is?

Question

Answer

Calculation of Magnetic Induction due to Current-Carrying Conducting Planes

Given:

Two large conducting planes carrying current perpendicular to x-axis

The planes are placed at (b, 0) and (20, 0)

Current per unit within both the planes is cm

The current is flowing in outward direction

Calculation:

Let's consider a point P at a distance x from the plane at (b, 0)

The magnetic field due to the current-carrying plane at (b, 0) can be calculated using Ampere's Law

It is given by B = μ₀I/2πx, where μ₀ is the permeability of free space, I is the current per unit area, and x is the distance from the plane

Now, let's consider a point Q at a distance x from the plane at (20, 0)

Similarly, the magnetic field due to the current-carrying plane at (20, 0) can be calculated using Ampere's Law

It is given by B = μ₀I/2π(x-20), where μ₀ is the permeability of free space, I is the current per unit area, and (x-20) is the distance from the plane

Since the two planes are carrying current in the same direction, the magnetic fields at point P and Q will add up

So, the total magnetic field at point P and Q will be B = μ₀I/2πx + μ₀I/2π(x-20)

Simplifying the equation, we get B = μ₀I/2π[(x(x-20))]

Conclusion:

The variation of magnetic induction as a function of x is given by B = μ₀I/2π[(x(x-20))]

As x increases, the magnetic induction decreases

The magnetic induction is maximum at x = 10, where it is equal to μ₀I/π

Answer

Sry but figure is missing

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