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A solenoid that is 97.9 cm long has a radius of 2.81 cm and a winding of 1370 turns; it carries a current of 3.17 A.

Calculate the magnitude of the magnetic field inside the solenoid.

Provide your answer in appropriate units.

Answer :

The magnitude of the magnetic field inside the solenoid is approximately 1.743 × 10⁻⁴ Tesla.

To calculate the magnitude of the magnetic field inside the solenoid, we can use the formula:

B = (μ₀ × n × I) / L

Where:
B is the magnetic field
μ₀ is the permeability of free space, approximately 4π × 10⁻⁷ Tm/A
n is the number of turns per unit length
I is the current flowing through the solenoid
L is the length of the solenoid

First, let's find the number of turns per unit length:

n = N / L

Where:
N is the total number of turns
L is the length of the solenoid

Substituting the given values, we have:

n = 1370 / 97.9 cm

Since the length is given in centimeters, we need to convert it to meters:

n = 1370 / 0.979 m

Now, let's calculate the number of turns per unit length:

n = 1399.19 turns/m

Next, we can substitute the values into the formula to find the magnetic field:

B = (4π × 10⁻⁷ Tm/A) × (1399.19 turns/m) × (3.17 A) / (0.979 m)

Simplifying the equation, we have:

B = 1.743 × 10⁻⁴ T

Therefore, the magnitude of the magnetic field inside the solenoid is approximately 1.743 × 10⁻⁴ Tesla.

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