A long insulated copper wire is closely wound as a spiral of ' $N$ ' turns. The spiral has inner radius ' $a$ ' and outer radius ' $b$ '. The spiral lies in the $X-Y$ plane and a steady current ' $I$ ' flows through the wire. The $Z$-component of the magnetic field at the center of the spiral is
$\frac{\mu_0 N I}{2(b-a)} \ln \left(\frac{b}{a}\right)$
$\frac{\mu_0 N I}{2(b-a)} \ln \left(\frac{b+a}{b-a}\right)$
$\frac{\mu_0 N I}{2 b} \ln \left(\frac{b}{a}\right)$
$\frac{\mu_0 N I}{2 b} \ln \left(\frac{b+a}{b-a}\right)$
Two circular coils $X$ and $Y$, having equal number of turns, carry equal currents in the same sence and subtend same solid angle at point $O$. If the smaller coil $X$ is midway between $O$ and $Y$, and If we represent the magnetic induction due to bigger coil $Y$ at $O$ as $B_Y$ and that due to smaller coil $X$ at $O$ as $B_X$, then $\frac{{{B_Y}}}{{{B_X}}}$ is
A helium nucleus makes a full rotation in a circle of radius $0.8$ metre in two seconds. The value of the magnetic field $B$ at the centre of the circle will be
Figure shows a square loop $ABCD$ with edge length $a$. The resistance of the wire $ABC$ is $r$ and that of $ADC$ is $2r$. The value of magnetic field at the centre of the loop assuming uniform wire is
A current $I$ flowing through the loop as shown in the adjoining figure. The magnetic field at centre $O$ is
Find out magnetic field at point $O$ ?