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: Field of an infinite straight wire (Biot-Savart)

Step 1 (Biot-Savart law): The magnetic field due to a small current element \(I\,d\vec{l}\) at a point distant \(\vec{r}\) is
\[ dB=\frac{\mu_0}{4\pi}\,\frac{I\,dl\,\sin\theta}{r^{2}}, \]
where \(\theta\) is the angle between the element and the line joining it to the point.

Step 2 (Geometry): Take a long straight wire along the vertical, and point \(P\) at perpendicular distance \(a\) from the wire. Let the element \(dl\) be at distance \(l\) from the foot \(O\) of the perpendicular. Then
\[ r=\sqrt{a^{2}+l^{2}},\qquad \sin\theta=\frac{a}{r}=\frac{a}{\sqrt{a^{2}+l^{2}}}. \]
Step 3 (Field of the element): Substituting,
\[ dB=\frac{\mu_0 I}{4\pi}\,\frac{a\,dl}{(a^{2}+l^{2})^{3/2}}. \]
All elements give field into the page at \(P\) (same direction), so we add magnitudes.

Step 4 (Integrate over the whole wire): For an infinite wire, \(l\) runs from \(-\infty\) to \(+\infty\):
\[ B=\frac{\mu_0 I a}{4\pi}\int_{-\infty}^{+\infty}\frac{dl}{(a^{2}+l^{2})^{3/2}}. \]
Using the standard integral \(\displaystyle\int_{-\infty}^{+\infty}\frac{dl}{(a^{2}+l^{2})^{3/2}}=\left[\frac{l}{a^{2}\sqrt{a^{2}+l^{2}}}\right]_{-\infty}^{+\infty}=\frac{2}{a^{2}}. \)

Step 5 (Result):
\[ B=\frac{\mu_0 I a}{4\pi}\cdot\frac{2}{a^{2}}=\frac{\mu_0 I}{2\pi a}. \]
The field lines are concentric circles around the wire; direction is given by the right-hand thumb rule.
\[ \boxed{\;B=\dfrac{\mu_0 I}{2\pi a}\;} \]

Option 2: Self vs mutual induction and the numerical

Step 1 (Self-induction): When the current in a coil changes, the flux linked with the same coil changes and an e.m.f. is induced in it opposing the change. This property is self-induction; \(e=-L\dfrac{dI}{dt}\), where \(L\) is the self-inductance of that coil.

Step 2 (Mutual induction): When the current in one coil (primary) changes, the flux linked with a neighbouring coil (secondary) changes and an e.m.f. is induced in the second coil; \(e_2=-M\dfrac{dI_1}{dt}\), where \(M\) is the mutual inductance. Key difference: self-induction involves a single coil, mutual induction involves two coupled coils.

Step 3 (Formula for the problem): Magnitude of induced e.m.f. in secondary:
\[ e=M\,\frac{dI}{dt}=M\,\frac{I-0}{t}. \]
Step 4 (Substitute): Given \(e=1500\text{ V}\), \(M=0.5\text{ H}\), \(t=10^{-3}\text{ s}\), final current \(=0\):
\[ 1500=0.5\times\frac{I}{10^{-3}}. \]
Step 5 (Solve for I):
\[ I=\frac{1500\times10^{-3}}{0.5}=\frac{1.5}{0.5}=3\text{ A}. \]
\[ \boxed{\;I=3\text{ A}\;} \]