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 at the centre of a circular coil is given by:

$B = \frac{\mu_0 N I}{2R}$,

where $N$ is the number of turns, 
$I$ is the current, 
$R$ is the radius, and 
$\mu_0$ is the permeability of free space. 
Substituting the values:

$B = \frac{(4\pi \times 10^{-7}) \times 100 \times 7}{2 \times 0.1} = 4.4 \, mT$.

Answer 2

Step 1: Use the formula for the magnetic field at the center of a circular coil. - The magnetic field at the center of a circular coil is given by: \[ B = \frac{\mu_0 N I}{2R} \] where N = 100 (number of turns), I = 7A (current), R = 0.1m (radius), and \(\mu_0 = 4\pi \times 10^{-7} Tm/A\) (permeability of free space). Step 2: Substitute the values into the formula. \[ B = \frac{(4\pi \times 10^{-7}) \cdot 100 \cdot 7}{2 \times 0.1} \] Simplify: \[ B = \frac{28\pi \times 10^{-7}}{0.2} \] \[ B = 4.4 \times 10^{-3} T = \mathbf{4.4mT} \]