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A (1.03x10^2)-mA current is used to charge up a parallel plate capacitor. A large square piece of paper is placed between the plates and parallel to them so it sticks out on all sides. What is the value of the integral ∮B⃗ .ds⃗ around the perimeter of the paper? Express your result in T.m with three significant figures.

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oladayoademosu

Since a 1.03 × 10² mA current is used to charge up the parallel plate capacitor and a large square piece of paper is placed between the plates and parallel to them so it sticks out on all sides, the value of ∫B.ds around the perimeter of the paper is 1.29 × 10⁻⁷ Tm

Ampere's Law

This shows the relationship between the magnetic field, B, the path of integration, ds and the current enclosed by the magnetic field, i.

Ampere's law is given mathematically as

∫B.ds = μ₀i where

  • B = magnetic field,
  • ds = arc length of path of integration,
  • μ₀ = permeability of free space = 4π × 10⁻⁷ Tm/A and
  • i = current enclosed by the path of integration, ds.

Given that for the large square piece of paper, the current used to charge the capacitor is

  • i = 1.03 × 10² mA = 1.03 × 10² × 10⁻³ A = 1.03 × 10⁻¹ A and
  • ds = path length around the perimeter of the paper

The value of ∫B.ds around the perimeter of the paper

Since ∫B.ds = μ₀i

Substituting the values of the variables into the equation, we have

∫B.ds = μ₀i

∫B.ds = 4π × 10⁻⁷ Tm/A × 1.03 × 10⁻¹ A

∫B.ds = 4.12π × 10⁻⁸ Tm

∫B.ds = 12.94 × 10⁻⁸ Tm

∫B.ds = 1.294 × 10⁻⁷ Tm

∫B.ds ≅ 1.29 × 10⁻⁷ Tm

So, the value of ∫B.ds around the perimeter of the paper is 1.29 × 10⁻⁷ Tm

Learn more about Ampere's law here:

https://brainly.com/question/17070619

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