Consider a tightly wound $100$ turn coil of radius $10 \; cm$,carrying a current of $1 \; A$. What is the magnitude of the magnetic field at the centre of the coil?

$A$ horizontal overhead powerline is at a height of $4\ m$ from the ground and carries a current of $100\ A$ from east to west. The magnetic field directly below it on the ground is: (Given: $\mu_0 = 4\pi \times 10^{-7} \, T \cdot m/A$)

$A$ long conducting wire having a current $I$ flowing through it is bent into a circular coil of $N$ turns. Then it is bent into a circular coil of $n$ turns. The magnetic field is calculated at the centre of the coils in both cases. The ratio of the magnetic field in the first case to that of the second case is:

Derive an expression for the force per unit length between two infinitely long straight parallel current-carrying wires. Hence,define one ampere $(A)$.

The figure shows a current-carrying square loop $\text{ABCD}$ of edge length '$a$' lying in a plane. If the resistance of the $\text{ABC}$ part is $r$ and that of the $\text{ADC}$ part is $2r$,then the magnitude of the resultant magnetic field at the centre of the square loop is:

The magnetic field at a perpendicular distance of $1 \,m$ from a long straight wire carrying a current of $1 \,A$ is: