Question

Option 1: With the help of the given figure, prove that the magnetic field produced at point P is \(B=\dfrac{\mu_0 i}{2d}\left[1+\dfrac{2}{\pi}\right]\). (The wire is hairpin-shaped: a semicircular arc joined to two long straight parallel segments carrying current \(i\); P is at the centre of the arc and the straight segments are a distance \(d\) apart.)
OR
Option 2: What is a radial magnetic field? Explain the working of a moving coil galvanometer with a suitable diagram. How can a galvanometer be converted into an ammeter and a voltmeter?

Hint:

Split the hairpin into a semicircular arc (field \(\mu_0 i/4R\)) and two semi-infinite straight wires (each \(\mu_0 i/4\pi a\)), with \(R=a=d/2\); add them in the same direction and factor out \(\mu_0 i/2d\).

Answer 1

Option 1: Magnetic field of the hairpin wire at P

Step 1: Identify the geometry. The hairpin has three parts: a semicircular arc and two long straight parallel wires. The two straight wires are separated by a distance \(d\), and P is the centre of the semicircle. Therefore the radius of the semicircle is
\[R=\frac{d}{2}.\]Each straight wire is at perpendicular distance \(a=R=\dfrac{d}{2}\) from P, and P lies on the line through the end of each straight wire (each straight part is 'semi-infinite', running from the arc to infinity).
Step 2: Field due to the semicircular arc. For a full circular loop the field at the centre is \(\mu_0 i/2R\); a semicircle is half of it:
\[B_{arc}=\frac{\mu_0 i}{4R}=\frac{\mu_0 i}{4\left(\dfrac{d}{2}\right)}=\frac{\mu_0 i}{2d}.\]
Step 3: Field due to one semi-infinite straight wire. For a straight wire the Biot-Savart result is \(B=\dfrac{\mu_0 i}{4\pi a}(\sin\theta_1+\sin\theta_2)\). For a semi-infinite wire whose near end is level with P, the angles are \(\theta_1=0^{\circ}\) (at the near end) and \(\theta_2=90^{\circ}\) (towards infinity):
\[B_{straight}=\frac{\mu_0 i}{4\pi a}(\sin0^{\circ}+\sin90^{\circ})=\frac{\mu_0 i}{4\pi a}=\frac{\mu_0 i}{4\pi\left(\dfrac{d}{2}\right)}=\frac{\mu_0 i}{2\pi d}.\]
Step 4: Both straight wires add. By the direction of current (Right Hand Rule), both straight segments produce a field at P in the same direction as the arc's field (into or out of the plane). So the two straight wires together give
\[B_{2\,straight}=2\times\frac{\mu_0 i}{2\pi d}=\frac{\mu_0 i}{\pi d}.\]
Step 5: Total field at P.
\[B=B_{arc}+B_{2\,straight}=\frac{\mu_0 i}{2d}+\frac{\mu_0 i}{\pi d}.\]Take \(\dfrac{\mu_0 i}{2d}\) common; note \(\dfrac{\mu_0 i}{\pi d}=\dfrac{\mu_0 i}{2d}\cdot\dfrac{2}{\pi}\):
\[B=\frac{\mu_0 i}{2d}\left(1+\frac{2}{\pi}\right).\]
\[\boxed{B=\frac{\mu_0 i}{2d}\left[1+\frac{2}{\pi}\right]}\quad\text{(proved)}\]

Option 2: Radial field and moving coil galvanometer

Radial magnetic field: A radial magnetic field is one in which the field lines are always directed along the radius, so that the plane of the current-carrying coil is always parallel to the field (and the coil's normal is always perpendicular to \(\vec{B}\)). It is produced by using concave (cylindrically curved) pole pieces of the magnet together with a soft-iron cylindrical core inside the coil.

Working of moving coil galvanometer (MCG): A rectangular coil of \(N\) turns and area \(A\) is suspended (or pivoted on a spring) in a radial magnetic field \(B\). When current \(I\) passes through it, the coil experiences a torque
\[\tau=NBIA\sin\theta.\]Because the field is radial, the plane of the coil is always parallel to \(\vec{B}\), so \(\theta=90^{\circ}\) and \(\sin\theta=1\) at every position. Hence the deflecting torque is \(NBIA\), independent of the deflection.
The suspension provides a restoring torque \(k\phi\) (where \(k\) is the torsion constant and \(\phi\) the deflection). At equilibrium:
\[NBIA=k\phi\ \Rightarrow\ I=\frac{k}{NBA}\,\phi.\]Thus \(I\propto\phi\): deflection is directly proportional to current, giving a uniform (linear) scale. This linearity is the main advantage of using a radial field.

Conversion into an ammeter: Connect a small resistance (shunt) \(S\) in parallel with the galvanometer of resistance \(G\). If \(I_g\) is the full-scale galvanometer current and \(I\) the maximum current to be measured,
\[S=\frac{I_g\,G}{I-I_g}.\]The low combined resistance keeps the ammeter's resistance small, so it is connected in series in the circuit.

Conversion into a voltmeter: Connect a high resistance \(R\) in series with the galvanometer. To read up to a maximum voltage \(V\),
\[R=\frac{V}{I_g}-G.\]The high total resistance makes the voltmeter draw very little current, so it is connected in parallel across the element whose voltage is measured.
\[\boxed{S=\frac{I_gG}{I-I_g}\ (\text{ammeter}),\qquad R=\frac{V}{I_g}-G\ (\text{voltmeter})}\]