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 4π × 10^–7 SI units):

The correct option is 4.4 mT.

To find the magnitude of the magnetic field at the centre of a circular coil, we use the formula for the magnetic field produced by a flat circular coil of $N$ turns:

$B=\frac{{\mu }_{0}NI}{2r}$

Where:
• $B$ is the magnetic field at the centre.
• ${\mu }_0$ is the permeability of free space, given as $4\pi \times {10}^{-7}\text{T}·\text{m/A}$.
• $N$ is the number of turns, which is $100$.
• $I$ is the current carrying through the coil, which is $7\text{A}$.
• $r$ is the radius of the coil, which is $10\text{cm}=0.1\text{m}$ (or ${10}^{-1}\text{m}$).

Now, let's substitute these values into the formula:

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

Using the approximation $\pi \approx \frac{22}{7}$:

$B=\frac{4\times \frac{22}{7}\times {10}^{-7}\times {10}^2\times 7}{0.2}$

Simplifying the numerator (canceling out the $7$ in the denominator of the fraction and the factor $7$):

$B=\frac{88\times {10}^{-5}}{0.2}$

Dividing by $0.2$ (which is equivalent to multiplying by $5$):

$B=440\times {10}^{-5}\text{T}$

Converting this to millitesla ($\text{mT}$), where $1\text{mT}={10}^{-3}\text{T}$:

$B=4.4\times {10}^{-3}\text{T}=4.4\text{mT}$

Thus, the magnitude of the magnetic field at the centre of the coil is 4.4 mT.