Q.A 100-turn coil of radius $1{\cdot}6$ cm and resistance $5{\cdot}0\ \Omega$ is co-axial with a solenoid of 250 turns/cm and radius $1{\cdot}8$ cm. The solenoid current drops from $1{\cdot}5$ A to zero in 25 ms. Calculate the current induced in the coil in this duration. (Take $\pi^2 = 10$)

The key idea is Mutual Inductance: the changing solenoid current changes the magnetic flux through the co-axial coil, inducing an EMF that drives a current through the coil's resistance.

1. The field inside the solenoid is $B_s = \mu_0 n I_s$. Since the coil (radius $r_c = 1{\cdot}6$ cm) sits entirely inside the solenoid's field region ($r_s = 1{\cdot}8$ cm), the flux linked with its $N_c$ turns is $N_c \Phi = N_c (\mu_0 n I_s)(\pi r_c^2)$, so the mutual inductance is $M = \mu_0 n N_c \pi r_c^2$.

2. With $N_c = 100$, $n = 250$ turns/cm $= 2{\cdot}5 \times 10^4$ m$^{-1}$, $r_c = 0{\cdot}016$ m and $\pi^2 = 10$:

$M = 40 \times 0{\cdot}25 \times 2{\cdot}56 \times 10^{-4} = 2{\cdot}56 \times 10^{-3}\ \text{H}$ …

When a changing current flows through one coil, it produces a changing magnetic field. If another coil is placed nearby, this changing magnetic field will pass through it, causing the magnetic flux linked with the second coil to change. According to Faraday's Law of Electromagnetic Induction, a changing magnetic flux through a coil induces an electromotive force (EMF) in that coil. If the second coil forms a closed circuit, this induced EMF will drive an induced current through its resistance. This phenomenon is known as mutual induction.

The extent to which a changing current in one coil induces an EMF in another is quantified by a property called mutual inductance ($M$). For a given pair of coils, $M$ depends on their geometry, number of turns, and relative orientation. Once $M$ is known, the induced EMF can be directly calculated from the rate of change of current in the primary coil.

Here, the solenoid acts as the primary coil, and the 100-turn coil acts as the secondary coil. As the current in the solenoid drops, the magnetic field it produces changes, inducing an EMF and subsequently a current in the co-axial coil.

1. Identify the given parameters:

   We are given the following information:

   • Number of turns in the coil, $N_c = 100$
   • Radius of the coil, $r_c = 1.6 \text{ cm} = 1.6 \times 10^{-2} \text{ m}$
   • Resistance of the coil, $R_c = 5.0 \ \Omega$
   • Number of turns per unit length of the solenoid, $n_s = 250 \text{ turns/cm} = 250 \times 100 \text{ turns/m} = 2.5 \times 10^4 \text{ turns/m}$
   • Radius of the solenoid, $r_s = 1.8 \text{ cm} = 1.8 \times 10^{-2} \text{ m}$
   • Initial current in the solenoid, $I_1 = 1.5 \text{ A}$
   • Final current in the solenoid, $I_2 = 0 \text{ A}$
   • Time duration for current change, $\Delta t = 25 \text{ ms} = 25 \times 10^{-3} \text{ s}$
   • Permeability of free space, $\mu_0 = 4\pi \times 10^{-7} \text{ T m/A}$
   • Given approximation, $\pi^2 = 10$

   We need to calculate the current induced in the coil.

2. Calculate the mutual inductance ($M$) between the solenoid and the coil:

   The magnetic field inside a long solenoid is uniform and given by $B_s = \mu_0 n_s I_s$, where $I_s$ is the current in the solenoid.

   Since the coil is co-axial with the solenoid and its radius ($r_c = 1.6 \text{ cm}$) is smaller than the solenoid's radius ($r_s = 1.8 \text{ cm}$), the magnetic field produced by the solenoid passes entirely through the area of the coil.

   The area of one turn of the coil is $A_c = \pi r_c^2$.

   The magnetic flux through one turn of the coil due to the solenoid's current is $\Phi_c = B_s A_c = (\mu_0 n_s I_s) (\pi r_c^2)$.

   The total magnetic flux linkage with the coil (which has $N_c$ turns) is $N_c \Phi_c = N_c (\mu_0 n_s I_s \pi r_c^2)$.

   The mutual inductance $M$ between a long solenoid and a co-axial coil placed inside it is given by:

   $M = \frac{N_c \Phi_c}{I_s} = \mu_0 n_s N_c A_c = \mu_0 n_s N_c (\pi r_c^2)$

   Now, substitute the given values into the formula for $M$:

   $M = (4\pi \times 10^{-7} \text{ T m/A}) \times (2.5 \times 10^4 \text{ turns/m}) \times (100 \text{ turns}) \times (\pi (1.6 \times 10^{-2} \text{ m})^2)$

   $M = (4\pi \times 10^{-7}) \times (2.5 \times 10^4) \times 100 \times (\pi \times 2.56 \times 10^{-4})$ …