A long straight wire, carrying current $I$ is bent at its mid-point to form an angle of $45^{\circ}$. Induction of magnetic field (in tesla) at point $P$, distant $R$ from point of bending is equal to
$\frac{(\sqrt{2}-1) \mu_{0} l}{4 \pi R}$
$\frac{(\sqrt{2}+1) \mu_{0} I}{4 \pi R}$
$\frac{(\sqrt{2}-1) \mu_{0} I}{4 \sqrt{2} \pi R}$
$\frac{(\sqrt{2}+1) \mu_{0} I}{4 \sqrt{2} \pi R}$
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 wire in the form of a circular loop of one turn carrying a current produces a magnetic field $B$ at the centre. If the same wire is looped into a coil of two turns and carries the same current, the new value of magnetic induction at the centre is
A cell is connected between two points of a uniformly thick circular conductor. The magnetic field at the centre of the loop will be
The magnetic field at the centre of a circular current carrying-conductor of radius $r$ is $B_c$. The magnetic field on its axis at a distance $r$ from the centre is $B_a$. The value of $B_c$ : $B_a$ will be
Which is a vector quantity