If we double the radius of a coil keeping the current through it unchanged,then the magnetic field at any point at a large distance from the centre becomes approximately

$A$ wire has three different sections as shown in the figure. The magnitude of the magnetic field produced at the centre '$O$' of the semicircle by the three sections together is $(\mu_0 = \text{permeability of free space})$:

In the Biot-Savart law,the direction of the magnetic field is determined by which of the following cross products in the expression $d\vec B = \frac{\mu_0}{4\pi} \frac{I d\vec l \times \vec r}{r^3}$?

The magnetic field at the center of a circular coil carrying current '$I$' for a single turn of a given length of wire is '$B$'. The same wire is bent into a circular coil having two turns. When the same current '$I$' passes through it, the value of the magnetic field becomes:

Two identical circular loops $P$ and $Q$ each of radius $r$ are lying in parallel planes such that they have a common axis. The currents through $P$ and $Q$ are $I$ and $4I$ respectively in the clockwise direction as seen from $O$. The net magnetic field at $O$ is:

$A$ telephone cable at a place has four long straight horizontal wires carrying a current of $1.0 \; A$ in the same direction from east to west. The earth's magnetic field at the place is $0.39 \; G$ and the angle of dip is $35^{\circ}$. The magnetic declination is nearly zero. What are the resultant magnetic fields at points $4.0 \; cm$ below and above the cable?