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 determine the current required to produce a magnetic field of 20 mT inside a long solenoid, we can use the formula for the magnetic field inside a solenoid: 

B = μ₀ n I

where:

• B is the magnetic field (20 mT or 0.02 T)
• μ₀ is the permeability of free space (4π x 10⁻⁷ T·m/A)
• n is the number of turns per meter
• I is the current in amperes

Since the solenoid is wound at 20 turns per cm, we convert this to turns per meter:

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

Substituting these values into the formula, we get:

0.02 = 4π x 10⁻⁷ x 2000 x I

Solving for I gives:

I = 0.02 / (4π x 10⁻⁷ x 2000)

I = 8 A

Thus, the required current is 8.0 A.

Answer 2

• Turn density: 20 turns/cm = 2000 turns/m
• Magnetic field: 20 mT = 0.02 T

Magnetic Field in a Solenoid:

\[ B = \mu_0 n I \]

where:
\( \mu_0 = 4\pi \times 10^{-7} \) Tm/A (permeability of free space)
\( n \) = number of turns per unit length
\( I \) = current

Calculations:

\[ 0.02 = (4\pi \times 10^{-7}) \times 2000 \times I \]

\[ I = \frac{0.02}{4\pi \times 10^{-7} \times 2000} \approx 7.96 \text{ A} \]

Rounding to nearest option: 8.0 A