Question

Option 1: Depict the series and the parallel combination of cells. When is the series and when is the parallel combination suitable? A cell of e.m.f. \( E \) and internal resistance \( r \) is connected with an external resistance \( R \). Draw a graph between the voltage \( V \) across \( R \) and the current \( i \) flowing in it.
OR
Option 2: State Ampere's circuital law and, on its basis, find the formula for the magnetic field produced at a point due to an infinitely long straight current-carrying wire. A current of \( 10^{-4}\ \text{A} \) flows in a rectangular coil of dimensions \( 1.0\ \text{cm} \times 1.5\ \text{cm} \). The plane of the coil is parallel to a magnetic field of \( 0.6\ \text{N/A m} \). The coil has 30 turns. Calculate the torque acting on the coil.

Hint:

Series cells: \(i=nE/(R+nr)\), good when \(R\gg r\); parallel: \(i=mE/(mR+r)\), good when \(R\ll r\); \(V=E-ir\) is a falling straight line. For the coil, \(B=\mu_0 I/2\pi r\) and, with plane parallel to \(B\), \(\tau=NIAB=2.7\times10^{-7}\) N m.

Answer 1

OPTION 1 (Cells - reasoning through equivalent source)

Step 1: Treat any group of cells as one equivalent battery of e.m.f. \(E_{eq}\) and internal resistance \(r_{eq}\). Series: e.m.f.s add and internal resistances add, so \(E_{eq}=nE\), \(r_{eq}=nr\); the current is \(i = \dfrac{nE}{R+nr}\).
Step 2: Series pays off only when the load dominates the internal resistance (\(R\gg r\)), otherwise the growing \(nr\) eats up the extra e.m.f. So use series for high-resistance external circuits.
Step 3: Parallel: the e.m.f. stays \(E\) but \(m\) equal internal resistances in parallel give \(r_{eq}=r/m\); current \(i = \dfrac{E}{R+r/m}\). This helps only when \(R\ll r\), so use parallel for low-resistance external circuits, where it multiplies the deliverable current.
Step 4: For one cell driving \(R\): current \(i=E/(R+r)\) and terminal p.d. \(V = iR = E - ir\). Since \(V\) falls by \(r\) for every unit rise in \(i\), plotting \(V\) (y) against \(i\) (x) gives a straight line starting at \(V=E\) (no current) and reaching \(V=0\) at \(i=E/r\). \[ \boxed{V = E - ir\ \text{(line: y-intercept } E,\ \text{x-intercept } E/r)} \]

OPTION 2 (Ampere's law and torque - alternate wording)

Step 1: Ampere's law: for a steady current, \(\oint \vec{B}\cdot d\vec{l} = \mu_0 I_{enc}\), the loop integral of \(B\) equalling \(\mu_0\) times the enclosed current.
Step 2: Around a long straight wire the field has the same magnitude everywhere on a coaxial circle of radius \(r\) and is directed along it, so the integral is simply \(B\times\) (circumference) \(= B(2\pi r)\). Setting this equal to \(\mu_0 I\) gives \(\boxed{B = \dfrac{\mu_0 I}{2\pi r}}\).
Step 3: A current loop is a magnetic dipole of moment \(\mu = NIA\). Its torque in a field is \(\vec{\tau} = \vec{\mu}\times\vec{B}\), magnitude \(NIAB\sin\theta\). With the coil plane along \(B\), the moment \(\vec{\mu}\) is perpendicular to \(B\), so \(\theta=90^\circ\) and \(\tau\) is maximum \(=NIAB\).
Step 4: Area \(A = (1.0\times10^{-2}\,\text{m})(1.5\times10^{-2}\,\text{m}) = 1.5\times10^{-4}\,\text{m}^2\). Then \(\mu = NIA = 30\times10^{-4}\times1.5\times10^{-4} = 4.5\times10^{-7}\,\text{A m}^2\).
Step 5: \(\tau = \mu B = 4.5\times10^{-7}\times 0.6 = 2.7\times10^{-7}\) N m. \[ \boxed{\tau = 2.7\times10^{-7}\ \text{N m}} \]