A very large electric current can flow inside a tornado from the clouds to the ground. If the magnetic field is measured to be 1.62×10-8 T at a distance of 8.33 km from the tornado, then what is the magnitude of this current?
Suppose that the current is created the magnetic field, that is similar to the magnetic field of the very long wire with the current. The magnetic field in this case according to the Biot-Savart Law is
B=μₒ•I/2•π•r, then
I =2•π•B•r/ μₒ = 2•π•1.62•10^-8•8.33•10^3/4• π•10^-7 =675 A
To determine the magnitude of the current, we can make use of Ampere's law. Ampere's law states that the line integral of the magnetic field around a closed loop is equal to the product of the current passing through the loop and the permeability of free space (μ₀).
Mathematically, Ampere's law can be written as:
∮ B · dl = μ₀ * I
Where:
- ∮ B · dl represents the line integral of the magnetic field B over a closed loop
- μ₀ is the permeability of free space (μ₀ ≈ 4π × 10^-7 T·m/A)
- I is the current passing through the loop
In this scenario, we assume a circular loop around the tornado at a distance of 8.33 km (or 8,330 m) from the tornado.
Given that the magnetic field B is measured to be 1.62 × 10^-8 T, we can plug in these values to determine the current I:
∮ B · dl = μ₀ * I
Using the equation above, we have:
(1.62 × 10^-8 T) * (2π * 8,330 m) = (4π × 10^-7 T·m/A) * I
Simplifying the equation, we find:
(1.62 × 10^-8 T) * (2π * 8,330 m) = (4π × 10^-7 T·m/A) * I
I ≈ (1.62 × 10^-8 T * 2π * 8,330 m) / (4π × 10^-7 T·m/A)
After canceling out the units and simplifying the equation further, we get:
I = (1.62 × 8,330 m) / (2 * 10^-7 A)
Solving this equation, we find:
I ≈ 1.09 × 10^3 A
Therefore, the magnitude of the current flowing inside the tornado is approximately 1.09 × 10^3 Amperes (A).