Mutual Induction Questions in English

(D) The magnetic field $B$ produced by the square loop at its center is $B = \frac{\mu_0 I}{\pi} \frac{4 \sin(45^\circ)}{L} = \frac{2\sqrt{2}\mu_0 I}{\pi L}$.
Since the circular loop is very small $(L >> R)$, we assume the magnetic field $B$ is uniform over the area of the circular loop.
The magnetic flux $\phi$ linked with the circular loop of radius $R$ is $\phi = B \times A = B \times (\pi R^2)$.
Substituting the value of $B$, we get $\phi = \left( \frac{2\sqrt{2}\mu_0 I}{\pi L} \right) \times (\pi R^2) = \frac{2\sqrt{2}\mu_0 I R^2}{L}$.
The mutual inductance $M$ is defined as $M = \phi / I$.
Therefore, $M = \frac{2\sqrt{2}\mu_0 R^2}{L}$.

(C) The induced e.m.f. $\varepsilon$ in coil $Q$ due to the changing current in coil $P$ is given by the formula $\varepsilon = -M \frac{dI}{dt}$.
Given the current in coil $P$ is $I = I_0 \sin(\omega t)$.
Calculating the rate of change of current: $\frac{dI}{dt} = \frac{d}{dt}(I_0 \sin(\omega t)) = I_0 \omega \cos(\omega t)$.
Substituting this into the e.m.f. equation: $\varepsilon = -M (I_0 \omega \cos(\omega t)) = -M I_0 \omega \cos(\omega t)$.
The magnitude of the induced e.m.f. is $|\varepsilon| = M I_0 \omega |\cos(\omega t)|$.
The maximum value of $\cos(\omega t)$ is $1$.
Therefore, the maximum induced e.m.f. is $\varepsilon_{\text{max}} = M \omega I_0$.

(A) The induced e.m.f. $(e)$ in the secondary coil is given by the formula: $e = M \cdot \frac{dI}{dt}$, where $M$ is the coefficient of mutual induction, $dI$ is the change in current, and $dt$ is the time interval.
Given values are: $M = 3 \text{ H}$, $e = 4 \text{ kV} = 4000 \text{ V}$, and $dI = I_1 - I_2 = 7 \text{ A} - 2 \text{ A} = 5 \text{ A}$.
Substituting these values into the formula: $4000 = 3 \cdot \frac{5}{dt}$.
Rearranging for $dt$: $dt = \frac{3 \cdot 5}{4000} = \frac{15}{4000} \text{ s}$.
$dt = 0.00375 \text{ s} = 3.75 \times 10^{-3} \text{ s}$.

(A) The magnetic field $B$ at the center of a circular coil of radius $r_1$ carrying current $I$ is given by $B = \frac{\mu_0 I}{2r_1}$.
Since $r_2 \ll r_1$, the magnetic field produced by the larger coil is approximately uniform over the area of the smaller coil.
The magnetic flux $\phi$ linked with the smaller coil of radius $r_2$ is $\phi = B \cdot A = \left( \frac{\mu_0 I}{2r_1} \right) (\pi r_2^2)$.
The mutual inductance $M$ is defined as $M = \frac{\phi}{I}$.
Substituting the expression for $\phi$, we get $M = \frac{\mu_0 \pi r_2^2}{2r_1}$.

(D) The magnetic field $B$ at the center of a large circular coil of radius $R$ carrying current $I$ is given by $B = \frac{\mu_0 I}{2R}$.
Since $R \gg r$, the magnetic field $B$ is approximately uniform over the area of the smaller coil of radius $r$.
The magnetic flux $\phi$ linked with the smaller coil is $\phi = B \cdot A$, where $A = \pi r^2$ is the area of the smaller coil.
Therefore, $\phi = \left( \frac{\mu_0 I}{2R} \right) (\pi r^2) = \left( \frac{\mu_0 \pi r^2}{2R} \right) I$.
The mutual inductance $M$ is defined by the relation $\phi = MI$, so $M = \frac{\mu_0 \pi r^2}{2R}$.
Thus, $M \propto \frac{r^2}{R}$.

(C) The coefficient of coupling $k$ is defined by the formula $M = k \sqrt{L_1 L_2}$, where $M$ is the mutual inductance, and $L_1, L_2$ are the self-inductances of the two coils.
Given values are $L_1 = 75 \text{ mH}$, $L_2 = 48 \text{ mH}$, and $M = 37.2 \text{ mH}$.
Rearranging the formula for $k$, we get $k = \frac{M}{\sqrt{L_1 L_2}}$.
Substituting the values: $k = \frac{37.2}{\sqrt{75 \times 48}}$.
Calculating the product: $75 \times 48 = 3600$.
Taking the square root: $\sqrt{3600} = 60$.
Now, $k = \frac{37.2}{60} = 0.62$.
Thus, the coefficient of coupling is $0.62$.

(B) The magnetic field $B$ at the center of a circular coil of radius $r_1$ carrying current $I$ is given by $B = \frac{\mu_0 I}{2 r_1}$.
Since $r_1 > r_2$, we assume the smaller coil (radius $r_2$) is placed in the magnetic field produced by the larger coil (radius $r_1$).
The magnetic flux $\phi$ linked with the smaller coil is $\phi = B \cdot A_2$, where $A_2 = \pi r_2^2$ is the area of the smaller coil.
$\phi = \left( \frac{\mu_0 I}{2 r_1} \right) (\pi r_2^2) = \frac{\mu_0 \pi r_2^2}{2 r_1} I$.
The mutual inductance $M$ is defined by the relation $\phi = M I$.
Comparing the two equations, we get $M = \frac{\mu_0 \pi r_2^2}{2 r_1}$.

(B) The induced e.m.f. in the second coil is given by the formula $\varepsilon = -M \frac{dI}{dt}$.
Given, $M = 0.005 \text{ H}$, $I = I_0 \sin \omega t$, $I_0 = 10 \text{ A}$, and $\omega = 60\pi \text{ rad s}^{-1}$.
Substituting the expression for current: $\varepsilon = -M \frac{d}{dt}(I_0 \sin \omega t) = -M I_0 \omega \cos \omega t$.
The magnitude of the induced e.m.f. is $|\varepsilon| = M I_0 \omega |\cos \omega t|$.
The maximum value of e.m.f. $(\varepsilon_{max})$ occurs when $|\cos \omega t| = 1$.
Therefore, $\varepsilon_{max} = M I_0 \omega$.
Substituting the values: $\varepsilon_{max} = 0.005 \times 10 \times 60\pi$.
$\varepsilon_{max} = 0.05 \times 60\pi = 3\pi \text{ V}$.
Thus, the correct option is $B$.

(C) The formula for mutual inductance $M$ is given by $M = \frac{N_2 \phi_2}{I_1}$.
Given:
$I_1 = 6$ $A$
$\phi_2 = 12 \times 10^{-4}$ $Wb/turn$
$N_2 = 2000$ turns
Substituting the values:
$M = \frac{2000 \times 12 \times 10^{-4}}{6}$
$M = \frac{24000 \times 10^{-4}}{6}$
$M = \frac{2.4}{6} = 0.4$ $H$.

(D) The induced electromotive force $(EMF)$ in the secondary coil is given by $\varepsilon = M \frac{dI_p}{dt}$.
Given $I_p = 12 \sin(50\pi t)$, then $\frac{dI_p}{dt} = 12 \times 50\pi \cos(50\pi t) = 600\pi \cos(50\pi t)$.
The maximum induced $EMF$ is $\varepsilon_{max} = M \times (600\pi) = (\pi \times 10^{-3}) \times 600\pi = 600\pi^2 \times 10^{-3} \text{ V}$.
Using $\pi^2 = 10$, we get $\varepsilon_{max} = 600 \times 10 \times 10^{-3} = 6 \text{ V}$.
The impedance $Z$ of the secondary coil is $Z = \sqrt{R^2 + X_L^2}$, where $X_L = \omega L$.
Here $\omega = 50\pi \text{ rad/s}$ and $L = (60/\pi) \times 10^{-3} \text{ H}$.
$X_L = 50\pi \times \frac{60}{\pi} \times 10^{-3} = 3000 \times 10^{-3} = 3 \text{ } \Omega$.
$Z = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \text{ } \Omega$.
The maximum induced current $I_{s,max} = \frac{\varepsilon_{max}}{Z} = \frac{6}{5} = 1.2 \text{ A}$.