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:

Updated On: Mar 27, 2025
  • 7.0 A
  • 9.0 A
  • 8.0 A
  • 10.0 A
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The Correct Option is C

Solution and Explanation

  • 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

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