Correct Answer is Option (B)

KEY CONCEPT : Magnetic field at the center of a circular coil of radius $R$ carrying

current is $B=\frac{{\mu }_{0}i}{2R}$

Given: $n\times \left(2\pi r^{\prime }\right)=2\pi R$

$\Rightarrow n{r}^{\prime }=R...\left(1\right)$

$B^{\prime }=\frac{n.{\mu }_{0}i}{2r^{\prime }}...\left(2\right)$

from $\left(1\right)$ and $\left(2\right),$ $B^{\prime }=\frac{n{\mu }_{0}i.n}{2\pi R}=n^{2}B$

Correct Answer is Option (C)

The magnetic field at a point on the axis of a circular loop at a distance $x$ from center is,

$B=\frac{{\mu }_{0}i{a}^2}{2\left(x^2+a^2\right)3/2}$ $B^{\prime }=\frac{{\mu }_{0}i}{2a}$

$\therefore$ $B^{\prime }=\frac{B.{\left(x^2+a^2\right)}^{3/2}}{a^3}$

Put $x=4$ & $a=3\Rightarrow B^{\prime }=\frac{54\left(5^3\right)}{3\times 3\times 3}=250\mu T$

Correct Answer is Option (A)

KEY CONCEPT : We know that the magnetic field produced by a current carrying circular coil of radius $r$

at its center is $B=\frac{{\mu }_0}{4\pi }\frac{I}{r}\times 2\pi$

Here $B_A=\frac{{\mu }_0}{4\pi }\frac{I}{R}\times 2\pi$ and

$B_B=\frac{{\mu }_0}{4\pi }\frac{2I}{2R}\times 2\pi$

$\Rightarrow \frac{B_A}{B{}_B}=1$

Correct Answer is Option (C)

Magnetic field due to current in wire $1$ at point $P$ distant $r$ from the wire is



$B=\frac{{\mu }_0}{4\pi }\frac{i_1}{r}\left[\cos \theta +\cos \theta \right]$

$B=\frac{{\mu }_0}{2\pi }\frac{i_1\cos \theta }{r}$ (directed perpendicular to the plane of paper, inwards)

The force exerted due to this magnetic field on current element $i_{2}dl$ is

$dF=i_{2}dlB\sin {90}^{\circ }$

$\therefore$ $dF=i_{2}dl\left[\frac{{\mu }_0}{2\pi }\frac{i_1\cos \theta }{r}\right]$

$=\frac{{\mu }_0}{2\pi r}{i}_1{i}_{2}dl\cos \theta$