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: Hairpin field, piece-by-piece with directions

Step 1: Fix the radius. The two straight rails are a gap \(d\) apart and P sits midway on the semicircle joining them, so the arc radius is \(R=d/2\) and each rail is a perpendicular distance \(d/2\) from P.

Step 2: Arc contribution using the angle formula. A circular arc subtending angle \(\phi\) at its centre gives \(B=\dfrac{\mu_0 i}{4\pi R}\,\phi\). For a semicircle \(\phi=\pi\):
\(B_{arc}=\dfrac{\mu_0 i}{4\pi R}\cdot\pi=\dfrac{\mu_0 i}{4R}=\dfrac{\mu_0 i}{4(d/2)}=\dfrac{\mu_0 i}{2d}.\)

Step 3: One rail as half of an infinite wire. An infinite straight wire gives \(\dfrac{\mu_0 i}{2\pi a}\) at distance \(a\). A semi-infinite wire ending level with the field point contributes exactly half of that:
\(B_{rail}=\dfrac12\cdot\dfrac{\mu_0 i}{2\pi a}=\dfrac{\mu_0 i}{4\pi a}=\dfrac{\mu_0 i}{4\pi(d/2)}=\dfrac{\mu_0 i}{2\pi d}.\)

Step 4: Superpose. Curling the fingers of the right hand along the current shows the arc and both rails push the field the same way through P, so magnitudes add:
\(B=B_{arc}+2B_{rail}=\dfrac{\mu_0 i}{2d}+2\cdot\dfrac{\mu_0 i}{2\pi d}=\dfrac{\mu_0 i}{2d}+\dfrac{\mu_0 i}{\pi d}.\)

Step 5: Factor out.
\(B=\dfrac{\mu_0 i}{2d}\left(1+\dfrac{2}{\pi}\right).\)
\[\boxed{B=\frac{\mu_0 i}{2d}\left[1+\frac{2}{\pi}\right]}\]

Option 2: Radial field and galvanometer (concept-first framing)

Step 1: Meaning of radial field. Shaped concave magnet poles plus a soft-iron cylindrical core make the field lines emerge radially, like spokes of a wheel. As the coil turns, its plane always stays parallel to \(\vec B\), so the field 'follows' the coil.

Step 2: Why it matters. The torque on an \(N\)-turn coil of area \(A\) carrying current \(I\) is \(\tau=NBIA\sin\theta\). In a radial field \(\theta=90^\circ\) always, so \(\tau=NBIA\) is constant, not varying with deflection.

Step 3: Balance and read-out. The spring's restoring torque \(k\phi\) balances it: \(NBIA=k\phi\), giving \(I=\dfrac{k}{NBA}\phi\). Current is proportional to deflection, so the scale is evenly spaced (linear).

Step 4: Make an ammeter. Put a tiny shunt \(S\) in parallel so most current bypasses the coil: \(S=\dfrac{I_g G}{I-I_g}\). Net resistance is very low, so it goes in series without disturbing the circuit.

Step 5: Make a voltmeter. Put a large resistance \(R\) in series so only a trickle flows through the coil: \(R=\dfrac{V}{I_g}-G\). Net resistance is very high, so it goes in parallel across the component.
\[\boxed{I\propto\phi;\quad S=\frac{I_gG}{I-I_g},\quad R=\frac{V}{I_g}-G}\]