Question

A tightly wound \(100\) turns coil of radius \(10 cm\) carries a current of \( 7 A\). The magnitude of the magnetic field at the centre of the coil is (Take permeability of free space as \(4\pi×10^{-7}\) SI units):

• A. \(44mT\)
• B. \(4.4T\)
• C. \(4.4mT\)
• D. \(44 T\)

Answer 1

The Correct Option is C

The magnetic field strength at the center of a tightly wound coil is calculated using the formula for the magnetic field at the center of a circular wire loop:

\[B=\frac{\mu_0 n I}{2R}\]

where:

• \(B\) represents the magnetic field strength,
• \(\mu_0 = 4\pi \times 10^{-7} \, \text{T}\cdot \text{m/A}\) is the permeability of free space,
• \(n = 100\) is the number of turns in the coil,
• \(I = 7 \, \text{A}\) is the current flowing through the coil,
• \(R = 0.1 \, \text{m}\) is the radius of the coil.

Substituting the given values into the formula yields:

\[B = \frac{(4\pi \times 10^{-7}) \cdot 100 \cdot 7}{2 \cdot 0.1}\]

Performing the calculation within the parentheses first:

\[B = \frac{(4\pi \times 10^{-7}) \cdot 700}{0.2}\]

Simplifying further:

\[B = \frac{2800\pi \times 10^{-7}}{0.2}\]

\[B = 14000\pi \times 10^{-7}\]

Using the approximation \(\pi \approx 3.1416\):

\[B = 14000 \times 3.1416 \times 10^{-7}\]

\[B \approx 43998.4 \times 10^{-7}\]

Converting the magnetic field to milliTesla (1 T = 1000 mT):

\[B \approx 4.4 \, \text{mT}\]

Consequently, the magnetic field at the coil's center is approximately \(4.4 \, \text{mT}\).