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 in series and parallel; V-i graph)

Step 1: Series combination. The cells are joined so that the positive terminal of one connects to the negative terminal of the next (\( +\,|\,-\;+\,|\,-\dots \)). For \( n \) identical cells, each of e.m.f. \( E \) and internal resistance \( r \), the total e.m.f. is \( nE \) and total internal resistance \( nr \). Current through external \( R \): \[ i_{series} = \frac{nE}{R + nr} \]
Step 2: When series is suitable. If \( R \gg r \), then \( nr \) is negligible and \( i_{series}\approx \dfrac{nE}{R} \), i.e. the current is \( n \) times that of one cell. So the series combination is useful when the external resistance is much larger than the internal resistance.
Step 3: Parallel combination. Here all positive terminals join together and all negative terminals join together. For \( m \) identical cells, the net e.m.f. is still \( E \) but the net internal resistance is \( r/m \). Current: \[ i_{parallel} = \frac{E}{R + \dfrac{r}{m}} = \frac{mE}{mR + r} \]
Step 4: When parallel is suitable. If \( R \ll r \), then \( i_{parallel}\approx \dfrac{mE}{r} \), i.e. \( m \) times the single-cell current. So the parallel combination is useful when the external resistance is much smaller than the internal resistance.
Step 5: V-i relation for one cell. For a single cell of e.m.f. \( E \), internal resistance \( r \), connected to \( R \), the current is \( i = \dfrac{E}{R+r} \). The terminal voltage across \( R \) is \[ V = E - i\,r \]
Step 6: The graph. \( V = E - ir \) is a straight line: when \( i = 0 \) (open circuit), \( V = E \) (y-intercept); the slope is \( -r \); \( V \) becomes zero at the short-circuit current \( i = E/r \) (x-intercept). Thus the V-i graph is a descending straight line from \( (0,\,E) \) to \( (E/r,\,0) \). \[ \boxed{\,V = E - i\,r\ \ (\text{straight line, intercept } E,\ \text{slope } -r)\,} \]

OPTION 2 (Ampere's law and torque on a coil)

Step 1: Ampere's circuital law. The line integral of the magnetic field around any closed loop equals \( \mu_0 \) times the net current threading the loop: \[ \oint \vec{B}\cdot d\vec{l} = \mu_0\,I_{enclosed} \]
Step 2: Field of an infinite straight wire. By symmetry the field lines are circles concentric with the wire, and \( B \) is constant in magnitude on a circle of radius \( r \), tangent to it. Choosing such a circular Amperian loop, \[ \oint \vec{B}\cdot d\vec{l} = B\,(2\pi r) = \mu_0 I \]
Step 3: Result. \[ \boxed{\,B = \frac{\mu_0 I}{2\pi r}\,} \] directed tangentially (given by the right-hand rule).
Step 4: Torque on the coil - formula. For a coil of \( N \) turns, area \( A \), carrying current \( I \) in a field \( B \), the torque is \( \tau = NIAB\sin\theta \), where \( \theta \) is the angle between the field and the normal to the coil. When the plane of the coil is parallel to the field, the normal is perpendicular to \( B \), so \( \theta = 90^\circ \) and \( \sin\theta = 1 \): the torque is maximum, \[ \tau = NIAB \]
Step 5: Insert the data. \( N = 30 \), \( I = 10^{-4}\ \text{A} \), \( A = 1.0\ \text{cm}\times 1.5\ \text{cm} = (1.0\times10^{-2})(1.5\times10^{-2}) = 1.5\times10^{-4}\ \text{m}^2 \), \( B = 0.6\ \text{T} \).
Step 6: Compute. \[ \tau = 30\times(10^{-4})\times(1.5\times10^{-4})\times 0.6 \] \[ \tau = (30\times 1.5\times 0.6)\times10^{-8} = 27\times10^{-8} \] \[ \boxed{\,\tau = 2.7\times10^{-7}\ \text{N m}\,} \]