$A$ particle having a charge $50 \ e$ is revolving in a circular path of radius $0.4 \ m$ with $1 \ r.p.s.$ The magnetic field produced at the centre of the circle is $(\mu_0 = 4 \pi \times 10^{-7} \ SI \ units$ and $e = 1.6 \times 10^{-19} \ C)$.

$A$ 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 $(b)$?

If $B_1$ is the magnetic field induction at a point on the axis of a circular coil of radius $R$ situated at a distance $R \sqrt{3}$ and $B_2$ is the magnetic field at the centre of the coil,then the ratio of $\frac{B_1}{B_2}$ is equal to

Two concentric circular coils of $10$ turns each are situated in the same plane. Their radii are $20 \, cm$ and $40 \, cm$ and they carry respectively $0.2 \, A$ and $0.3 \, A$ current in opposite directions. The magnetic field at the centre is $(\mu_0 = 4 \pi \times 10^{-7} \, T \cdot m/A)$

Two concentric coplanar circular loops of radii $r_1$ and $r_2$ carry currents of respectively $i_1$ and $i_2$ in opposite directions (one clockwise and the other anticlockwise). The magnetic induction at the centre of the loops is half that due to $i_1$ alone at the centre. If $r_2 = 2r_1$,the value of $i_2/i_1$ is:

$A$ coil having $N$ turns is wound tightly in the form of a spiral with inner and outer radii $a$ and $b$ respectively. When a current $i$ passes through the coil,the magnetic field at the centre is