Two circular coils $1$ and $2$ are made from the same wire. The radius of the first coil is twice that of the second coil. What is the ratio of potential difference applied across them $V_1 / V_2$,so that the magnetic field at their centre is the same?

Consider a tightly wound $100$ turn coil of radius $10 \,cm$ carrying a current of $2 \,A$. The magnitude of the magnetic field at the centre of the coil is

For a circular coil of radius $R$ and $N$ turns carrying current $I$,the magnitude of the magnetic field at a point on its axis at a distance $x$ from its centre is given by,
$B=\frac{\mu_{0} I R^{2} N}{2\left(x^{2}+R^{2}\right)^{3 / 2}}$
$(a)$ Show that this reduces to the familiar result for field at the centre of the coil.
$(b)$ Consider two parallel co-axial circular coils of equal radius $R$ and number of turns $N,$ carrying equal currents in the same direction,and separated by a distance $R$. Show that the field on the axis around the mid-point between the coils is uniform over a distance that is small as compared to $R,$ and is given by,
$B=0.72 \frac{\mu_{0} N I}{R}, \quad \text { approximately }$

$A$ long straight wire carrying a current of $30\,A$ is placed in an external uniform magnetic field of induction $4 \times 10^{-4}\,T$. The magnetic field is acting parallel to the direction of the current. The magnitude of the resultant magnetic induction in tesla at a point $2.0\,cm$ away from the wire is

$A$ circular coil is in the $y-z$ plane with its centre at the origin. The coil is carrying a constant current. Assuming the direction of the magnetic field at $x = -25\, cm$ to be the positive direction of the magnetic field,which of the following graphs shows the variation of the magnetic field along the $x$-axis?

$A$ current-carrying wire in its neighbourhood produces: