The magnitude of magnetic field inside a solenoid of length 0.3 m having 800 turns carrying a current of 6 A is

The correct option is 20 mT.

To find the magnitude of the magnetic field inside a solenoid, we use the standard formula for the magnetic field of an ideal solenoid:
$$B={\mu }_0·n·I$$

where:
• $B$ is the magnetic field,
• ${\mu }_0$ is the permeability of free space, which is approximately equal to $4\pi \times {10}^{-7}\text{ T·m/A}$,
• $n$ is the number of turns per unit length ($n=\frac{N}{L}$),
• $I$ is the electric current flowing through the solenoid.

From the given data:
• Length of the solenoid, $L=0.3\text{ m}$,
• Total number of turns, $N=800$,
• Current, $I=6\text{ A}$.

First, calculate the number of turns per unit length ($n$):
$$n=\frac{800}{0.3}\approx 2666.67\text{ turns/m}$$

Now, substitute these values into the magnetic field formula:
$$B=(4\pi \times {10}^{-7}\text{ T·m/A})·\left(\frac{800}{0.3}\text{ m}{}^{-1}\right)·\left(6\text{ A}\right)$$

Simplifying the numerical expression:
$$B=4\pi \times {10}^{-7}\times \frac{4800}{0.3}$$
$$B=4\pi \times {10}^{-7}\times 16000$$
$$B=64000\pi \times {10}^{-7}\text{ T}$$

Substituting the value of $\pi \approx 3.14159$:
$$B\approx 64000\times 3.14159\times {10}^{-7}\text{ T}$$
$$B\approx 201061.9\times {10}^{-7}\text{ T}$$
$$B\approx 0.0201\text{ T}=20.1\text{ mT}$$

Rounding to the nearest options provided, we get:
$$B\approx 20\text{ mT}$$