The magnetic field at the centre of a circular coil of radius $r$ carrying current $I$ is $B_1$. The field at the centre of another coil of radius $2r$ carrying the same current $I$ is $B_2$. The ratio $\frac{B_1}{B_2}$ is

An infinitely long straight conductor is bent into the shape as shown in the figure. It carries a current of $i$ $ampere$ and the radius of the circular loop is $r$ $metre$. Then the magnetic induction at its centre $O$ will be:

An infinitely long wire carrying $1 \ A$ current in the $+z$ direction is placed at $(1 \ cm, 1 \ cm)$. Another wire carrying $1 \ A$ in the $+x$ direction is placed at $y=1 \ cm$. If the magnetic field due to this configuration at the origin is $B$. Let $B_0$ be the magnitude of the field if only the wire at $(1 \ cm, 1 \ cm)$ was present,then $\frac{B}{B_0}$ is

$A$ circular coil of wire consisting of $100$ turns each of radius $9 \ cm$ carries a current of $0.4 \ A$. The magnitude of the magnetic field at the centre of the coil is $[\mu_0 = 12.56 \times 10^{-7} \text{ SI Units}]$.

The Earth's magnetic induction at a certain point is $7 \times 10^{-5} \ Wb/m^2$. This is to be annulled by the magnetic induction at the center of a circular conducting loop of radius $5 \ cm$. The required current in the loop is......$A$

$A$ long straight wire carrying a current of $30 \ A$ is placed in an external uniform magnetic field of induction $4 \times 10^{-4} \ T$. The magnetic field is acting parallel to the direction of current. The magnitude of the resultant magnetic induction in tesla at a point $2.0 \ cm$ away from the wire is $(\mu_0 = 4 \pi \times 10^{-7} \ H/m)$.