Question

Two circular coils $\text{A}$ and $\text{B}$ of radii $a$ and $b$ respectively $(\text{with}b>>a)$ have their planes perpendicular to the plane of the page. They are separated co-axially by a distance $x=\sqrt{3}b$ as shown in the figure. A transient current $I$ flows through coil $\text{B}$ for a very short time interval. If the resistance of coil $\text{A}$ is $R$, then the charge that flows through coil $\text{A}$ during the short time interval will be

• A. $\frac{{\mu }_{0}I\pi a^2}{16bR}$
• B. $\frac{{\mu }_{0}I\pi a^2}{4bR}$
• C. $\frac{{\mu }_{0}I\pi a^2}{bR}$
• D. $\frac{{\mu }_{0}I\pi a^2}{2bR}$

Answer 1

The correct option is A $\frac{{\mu }_{0}I\pi a^2}{16bR}$

Magnetic field at the centre of coil $\text{A}$ due to current $I$ in coil $\text{B}$ is

$B_{AB}=\frac{{\mu }_{0}I{b}^2}{2(b^2+x^2{)}^{3/2}}$

$\because x=\sqrt{3}b$

$\therefore B_{AB}=\frac{{\mu }_{0}I{b}^2}{2(b^2+(\sqrt{3}b{)}^2{)}^{3/2}}$

$\Rightarrow B_{AB}=\frac{{\mu }_{0}I}{16b}$

Since, the magnetic field is along the axis of coil $\text{A}$, it is perpendicular to the plane of $\text{A}$, hence $\theta =0$.

Therefore, magnetic flux through $\text{A}$ is

$\varphi =B_{AB}\times \text{area of coil A}\times \cos 0^{\circ }$

$\Rightarrow \varphi =\frac{{\mu }_{0}I\pi a^2}{16b}$

Induced emf, $E=\left|\frac{d\varphi }{dt}\right|$

$\Rightarrow IR=\frac{d\varphi }{dt}$

$\Rightarrow IRdt=d\varphi$

$\Rightarrow \int Idt=\frac{1}{R}\int d\varphi$

$\Rightarrow Q=\frac{\varphi }{R}$

$\therefore Q=\frac{{\mu }_{0}I\pi a^2}{16bR}$

Hence, option $\left(\text{A}\right)$ is correct.