The magnetic field due to a current-carrying circular loop of radius $6 \ cm$ at a point on the axis at a distance of $8 \ cm$ from the centre is $216 \ \mu T$. Then the magnetic field at the centre of the ring is $\dots \ \mu T$.

$A$ straight wire carrying a current of $12\; A$ is bent into a semi-circular arc of radius $2.0\; cm$ as shown in Figure $(a)$. Consider the magnetic field $B$ at the centre of the arc.
$(a)$ What is the magnetic field due to the straight segments?
$(b)$ In what way does the contribution to $B$ from the semicircle differ from that of a circular loop and in what way does it resemble?
$(c)$ Would your answer be different if the wire were bent into a semi-circular arc of the same radius but in the opposite way as shown in Figure $(b)$?

Two circular coils $X$ (smaller) and $Y$ (bigger), each having a single turn, carry equal currents in the same direction and subtend the same angle at point '$O$' along the axis of the coil. The distances of the centers of coil to point '$O$' are $d$ and $(d/2)$ for coil $Y$ and $X$ respectively. The radii of coils $Y$ and $X$ are $(2r)$ and $(r)$ respectively. The magnetic induction due to the bigger coil at point '$O$' is $B_y$ and that due to smaller coil $X$ at point '$O$' is $B_x$. $(d \gg r)$ The relation between $B_x$ and $B_y$ is

On connecting a battery to the two corners of a diagonal of a square conductor frame of side $a$,the magnitude of the magnetic field at the centre will be:

$A$ particle having a charge $100 e$ is revolving in a circular path of radius $0.8 \ m$ with $1 \ r.p.s$. The magnetic field produced at the center of the circle in $SI$ unit is ($\mu_0$ is the permeability of vacuum,$e = 1.6 \times 10^{-19} \ C$).

$A$ direct current is passing through a wire. It is bent to form a coil of one turn. Now,it is further bent to form a coil of two turns but at a smaller radius. Find the ratio of the magnetic induction at the centre of this two-turn coil to that at the centre of the one-turn coil.