Part (i): Definition and SI Unit
Step 1 (Definition): [0.5 Mark]
Mutual inductance of a pair of coils is defined as the magnetic flux linked with one coil when a unit current flows through the other coil. Alternatively, it is the induced emf produced in one coil when the rate of change of current in the other coil is unity.
Step 2 (SI Unit): [0.5 Mark]
The SI unit of mutual inductance is Henry (H).
Part (ii): Derivation of Mutual Inductance
Step 1 (Governing Formula): [0.5 Mark]
The magnetic field $B$ inside a long solenoid carrying current $I$ is given by:
$$B = \mu_0 n I$$The magnetic flux $\phi$ linked with the small circular loop of radius $r$ is $\phi = B \cdot A$, where $A = \pi r^2$.
Step 2 (Mathematical Working): [1 Mark]
Since the loop is placed inside the solenoid, the magnetic field is uniform across the area of the loop. The flux linked with the loop is:
$$\phi = (\mu_0 n I) \cdot (\pi r^2)$$By definition of mutual inductance ($M$), the flux linked with the loop is $\phi = M I$. Equating the two expressions:
$$M I = \mu_0 n I \pi r^2$$Step 3 (Final Concluded Answer): [0.5 Mark]
Canceling $I$ from both sides, we get the mutual inductance:
$$M = \mu_0 n \pi r^2$$(Note: Since the loop has only one turn, the expression remains as above.)
💡 EL10 Concept Insight
Think of the solenoid as a "magnetic pipe" creating a steady field inside. When you place a loop inside, it "catches" some of that magnetic field; if you change the current in the pipe, the amount of field the loop catches changes, which creates electricity in the loop—that "catching efficiency" is exactly what we call Mutual Inductance!
OR: Force between Two Parallel Conductors
Step 1 (Governing Law & Given Data): [1 Mark]
The magnetic field $B_a$ produced by conductor A at a distance $d$ is:
$$B_a = \frac{\mu_0 I_a}{2 \pi d}$$The force $F$ on a current-carrying conductor of length $L$ in a magnetic field is $F = I L B \sin \theta$. Here, $\theta = 90^\circ$.
Step 2 (Mathematical Working): [1.5 Marks]
The force on conductor B due to the field of A is:
$$F_{ba} = I_b L B_a = I_b L \left( \frac{\mu_0 I_a}{2 \pi d} \right)$$ $$F_{ba} = \frac{\mu_0 I_a I_b L}{2 \pi d}$$Similarly, the force on conductor A due to the field of B ($B_b = \frac{\mu_0 I_b}{2 \pi d}$) is:
$$F_{ab} = I_a L B_b = \frac{\mu_0 I_b I_a L