Question

Derive the formula for magnetic field due to an infinitely long current carrying conducting wire using Biot-Savart law.
OR
Differentiate between self-induction and mutual induction. The current in primary coil becomes zero in \(10^{-3}\) s and an induced e.m.f. of 1500 volts is produced in secondary. Mutual inductance of coils is 0.5 H. Find the value of initial current in the primary.

Hint:

Integrate the Biot-Savart element \(dB=\tfrac{\mu_0}{4\pi}\tfrac{I\,dl\sin\theta}{r^2}\) over the whole wire to get \(B=\mu_0 I/2\pi a\). For the coil, \(e=M\,\Delta I/\Delta t\Rightarrow I=e\,\Delta t/M\).

Answer 1

Option 1: Straight-wire field via the angle form of Biot-Savart

Step 1: The Biot-Savart law states each element \(I\,dl\) contributes \(dB=\dfrac{\mu_0}{4\pi}\dfrac{I\,dl\sin\theta}{r^2}\) at a field point, directed perpendicular to the plane of \(dl\) and \(r\).

Step 2: Place point \(P\) at perpendicular distance \(a\). Instead of the length variable, use the angle \(\phi\) that \(r\) makes with the perpendicular. Then \(l=a\tan\phi\), \(dl=a\sec^2\phi\,d\phi\), and \(r=a\sec\phi\). Also the angle between element and \(r\) gives \(\sin\theta=\cos\phi\).

Step 3: Substituting, \(dB=\dfrac{\mu_0 I}{4\pi}\dfrac{(a\sec^2\phi\,d\phi)(\cos\phi)}{a^2\sec^2\phi}=\dfrac{\mu_0 I}{4\pi a}\cos\phi\,d\phi\). The awkward geometry collapses to a simple cosine.

Step 4: For an infinitely long wire the angle \(\phi\) sweeps from \(-90^\circ\) to \(+90^\circ\): \(B=\dfrac{\mu_0 I}{4\pi a}\displaystyle\int_{-\pi/2}^{+\pi/2}\cos\phi\,d\phi=\dfrac{\mu_0 I}{4\pi a}\,[\sin\phi]_{-\pi/2}^{+\pi/2}=\dfrac{\mu_0 I}{4\pi a}(1-(-1))\).

Step 5: This gives \(B=\dfrac{\mu_0 I}{2\pi a}\), matching Ampere's-law result, with circular field lines wrapping the wire (right-hand grip rule).
\[ \boxed{\;B=\dfrac{\mu_0 I}{2\pi a}\;} \]

Option 2: Comparing inductions and back-calculating the current

Step 1: Self-induction acts within one coil: a changing current makes its own flux change and it induces an opposing e.m.f. on itself, measured by self-inductance \(L\) (\(e=-L\,dI/dt\)).

Step 2: Mutual induction acts across two coils: a changing current in the primary alters the flux threading a nearby secondary, inducing an e.m.f. there, measured by mutual inductance \(M\) (\(e_2=-M\,dI_1/dt\)). So the distinction is one coil versus a pair of magnetically linked coils.

Step 3: Only the magnitude matters here. The secondary e.m.f. is \(e=M\left|\dfrac{\Delta I}{\Delta t}\right|\), and \(\Delta I=I_{\text{final}}-I_{\text{initial}}=0-I=-I\), so \(|\Delta I|=I\).

Step 4: Rearranging for the unknown initial current: \(I=\dfrac{e\,\Delta t}{M}\). Insert \(e=1500\) V, \(\Delta t=10^{-3}\) s, \(M=0.5\) H.

Step 5: \(I=\dfrac{1500\times10^{-3}}{0.5}=\dfrac{1.5}{0.5}=3\) A. The primary was carrying 3 A before it dropped to zero.
\[ \boxed{\;I=3\text{ A}\;} \]