Question

A closely wound coil of 100 turns and area of cross-section 2 cm^2 has a coefficient of self-induction 1 mH. The magnetic induction in the center of the core of the coil when a current of 2 A flows in it will be: _______

Answer 1

The magnetic induction in the center of the core of the coil when a current of 2 A flows in it is 4π * 10^-3 T.

Step-by-step explanation:

To find the magnetic induction in the center of the core of the coil when a current of 2 A flows in it, we can use the formula:

B = (μ₀ * N * I) / A

Where:

• B is the magnetic induction
• μ₀ is the permeability of free space (4π * 10^-7 T * m/A)
• N is the number of turns in the coil (100 turns)
• I is the current flowing through the coil (2 A)
• A is the area of cross-section of the coil (2 cm² = 2 * 10^-4 m²)

Substituting the values into the formula, we get:

B = (4π * 10^-7 T * m/A * 100 turns * 2 A) / (2 * 10^-4 m²)

Simplifying this expression gives us:

B = 4π * 10^-3 T

Therefore, the magnetic induction in the center of the core of the coil is 4π * 10^-3 T.