Question

A long solenoid is formed by winding insulated copper wire at the rate of 20 turns per cm. The current required to produce a magnetic field of 20 mT inside the solenoid at its center would be:

• A. 7.0 A
• B. 9.0 A
• C. 8.0 A
• D. 10.0 A

Answer 1

The Correct Option is C

To find the current needed for a magnetic field of 20 mT in a long solenoid, we use the formula:

B = μ₀ n I

Key variables are:

• B: Magnetic field strength (20 mT = 0.02 T)
• μ₀: Permeability of free space (4π x 10⁻⁷ T·m/A)
• n: Turns per meter
• I: Current in amperes

Given the winding density of 20 turns/cm, we convert to turns per meter:

n = 20 turns/cm * 100 cm/m = 2000 turns/m

Substituting the known values into the formula:

0.02 T = (4π x 10⁻⁷ T·m/A) * (2000 turns/m) * I

Solving for I:

I = 0.02 T / (4π x 10⁻⁷ T·m/A * 2000 turns/m)

I = 8 A

Therefore, the required current is 8.0 A.