Solution for Part (a): Force between two parallel current-carrying conductors
Step 1 (Governing Law & Setup): [1 Mark]
Consider two long, straight, parallel conductors $A$ and $B$ separated by a distance $d$, carrying currents $I_1$ and $I_2$ in opposite directions. According to the Right-Hand Thumb Rule, the magnetic field $B_1$ produced by conductor $A$ at the position of conductor $B$ is:
$$B_1 = \frac{\mu_0 I_1}{2\pi d}$$The direction of $B_1$ is perpendicular to the plane of the conductors (directed into the page). By Fleming’s Left-Hand Rule, the force on a segment of length $L$ of conductor $B$ is repulsive.
Step 2 (Mathematical Derivation): [1.5 Marks]
The magnetic force $F$ on a current-carrying conductor of length $L$ in a magnetic field $B$ is given by $F = I_2 L B_1 \sin(90^\circ)$. Substituting the value of $B_1$:
$$F = I_2 L \left( \frac{\mu_0 I_1}{2\pi d} \right)$$The force per unit length ($f = F/L$) is:
$$f = \frac{\mu_0 I_1 I_2}{2\pi d}$$Since the currents are in opposite directions, the force is repulsive.
Step 3 (Definition of One Ampere): [0.5 Mark]
One Ampere is defined as that constant current which, when maintained in each of the two infinitely long, straight, parallel conductors of negligible cross-section placed 1 meter apart in vacuum, produces a force of $2 \times 10^{-7} \, \text{N/m}$ between them.
💡 EL10 Concept Insight
Think of the magnetic fields around the wires like invisible gears; when currents flow in opposite directions, the magnetic fields "push" against each other in the space between the wires, forcing them apart just like two magnets with the same poles facing each other.
Solution for Part (b): Torque on a current loop in a uniform magnetic field
Step 1 (Governing Law & Diagram Description): [1 Mark]
Consider a rectangular loop $PQRS$ of sides $a$ and $b$ carrying current $I$ in a uniform magnetic field $\vec{B}$. The torque $\vec{\tau}$ is given by the cross product of the magnetic moment $\vec{m}$ and the magnetic field $\vec{B}$:
$$\vec{\tau} = \vec{m} \times \vec{B}$$(Diagram: A rectangular loop tilted at an angle $\theta$ with the magnetic field lines, showing forces $F_1$ and $F_2$ acting on the sides of length $a$ in opposite directions, creating a couple.)
Step 2 (Mathematical Derivation): [1.5 Marks]
The forces on the arms of length $b$ are equal and opposite, cancelling out. The forces on the arms of length $a$ are $F = I a B$. These forces form a couple with a perpendicular distance $b \sin\theta$ between them.
$$\tau = \text{Force} \times \text{perpendicular distance}$$ $$\tau = (I a B) \times (b \sin\theta)$$Since Area $A = a \times b$:
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