Question

A researcher would like to perform an experiment in a zero magnetic field, which means that the field of the earth must be canceled. Suppose the experiment is done inside a solenoid of diameter 1.0 m, length 3.0 m , with a total of 5000 turns of wire. The solenoid is oriented to produce a field that opposes and exactly cancels the 52 μT local value of the earth's field. What current is needed in the solenoid's wires?

Answer 1

Answer:

  I = 2.48 10⁻² A

Explanation:

The magnetic field produced by a solenoid disproportionate to the density of turns and the current passing through the wire

        B = μ₀ n I

Where  μ₀ is the magnetic permeability that you are worth 4π 10⁻⁷ N/A²

Let's calculate the density of turns.

        n = N / L

        n = 5000 / 3.0

        n = 1666.67

Let's reduce the field to SI units

      B = 52 μT = 52 10⁻⁶ T

Let's calculate the current

      I = B /  μ₀ n

      I = 52 10⁻⁶ / (4π 10⁻⁷ 1666.67)

      I = 2.48 10⁻² A

Answer 2

The current needed in the solenoid's wires is 2.48 10⁻² A.

What is Magnetic field?

The magnetic field which can be found in a solenoid is disproportionate to the density of turns and the current passing through it.

This formula is vital  B = μ₀ n I

where μ₀ is the magnetic permeability = 4π 10⁻⁷ N/A²

• Let's calculate the density of turns.

n = N / L

n = 5000 / 3.0

n = 1666.67

• Let's calculate the current.

We can convert to the standard unit

B = 52 μT = 52 × 10⁻⁶ T

I = B /  μ₀ n

I = 52 10⁻⁶ / (4π 10⁻⁷ 1666.67)

I = 2.48 10⁻² A.

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