Question

Explain Biot-Savart law. Deduce the formula for magnetic field produced due to a straight current carrying conductor.
OR
An α-particle, accelerated with the potential difference of V volt strikes with a nucleus (atomic number = Z). If r be the nearest distance of closest approach to the nucleus then prove that r = 14·4 Z/V Å.

Hint:

Biot-Savart: \(dB=\frac{\mu_0}{4\pi}\frac{I\,dl\sin\theta}{r^2}\); integrate along the wire to get \(B=\frac{\mu_0 I}{2\pi a}\). Closest approach: set kinetic energy \(2eV\) equal to Coulomb potential energy \(\frac{1}{4\pi\varepsilon_0}\frac{2Ze^2}{r}\) and solve for r.

Answer 1

OPTION 1: Biot-Savart Law and Straight Conductor

Step 1: What the law says. Biot-Savart law gives the magnetic contribution of one current element. For an element \(I\,d\vec{l}\) at position \(\vec{r}\) from an observation point, the field is
\[d\vec{B} = \frac{\mu_0}{4\pi}\frac{I\,d\vec{l}\times\hat{r}}{r^2},\qquad dB = \frac{\mu_0}{4\pi}\frac{I\,dl\sin\theta}{r^2}\]
with \(\theta\) the angle between \(d\vec{l}\) and \(\vec{r}\). The field is largest sideways to the element (\(\theta=90^\circ\)) and zero along its axis (\(\theta=0\)).

Step 2: Choice of variable. Take the straight wire along a line and let \(P\) be at perpendicular distance \(a\). Instead of \(\phi\), measure each element by the angle \(\alpha\) that the line from the element to \(P\) makes with the wire itself. Then \(\theta = \alpha\) and \(\sin\theta = \sin\alpha\).

Step 3: Relations. With \(O\) the foot of the perpendicular and \(l\) the distance of the element from \(O\), geometry gives \(r = a/\sin\alpha\) and \(l = -a\cot\alpha\), so \(dl = a\,\csc^2\alpha\,d\alpha\).

Step 4: Substitute.
\[dB = \frac{\mu_0 I}{4\pi}\frac{(a\csc^2\alpha\,d\alpha)\sin\alpha}{(a/\sin\alpha)^2} = \frac{\mu_0 I}{4\pi a}\sin\alpha\,d\alpha\]

Step 5: Integrate between the end angles. If the two ends of the wire correspond to angles \(\alpha_1\) and \(\alpha_2\) measured from the wire,
\[B = \frac{\mu_0 I}{4\pi a}\int_{\alpha_1}^{\alpha_2}\sin\alpha\,d\alpha = \frac{\mu_0 I}{4\pi a}\big(\cos\alpha_1 - \cos\alpha_2\big)\]
This is the same result as before written with angles from the wire.

Step 6: Long wire limit. For an infinite wire the ends go to \(\alpha_1 \to 0\) and \(\alpha_2 \to 180^\circ\), giving \(\cos 0 - \cos 180^\circ = 1-(-1)=2\). Hence
\[B = \frac{\mu_0 I}{4\pi a}\times 2 = \frac{\mu_0 I}{2\pi a}\]
the familiar field of a long straight wire, circling the conductor.
\[\boxed{B = \frac{\mu_0 I}{2\pi a}}\]

OPTION 2: Distance of Closest Approach of α-particle

Step 1: Work-energy view. The accelerating field does work \(W = qV\) on the α-particle. Its charge is twice the electron charge, \(q = 2e\), so it enters the target region with kinetic energy \(K = 2eV\) joule. No energy is lost on the way in, so this is the energy budget for the head-on collision.

Step 2: Turning point. The nucleus of charge \(Ze\) repels the positive α-particle. The particle slows, stops at separation \(r\), then returns. At that turning point its speed is zero, so kinetic energy has been fully traded for Coulomb potential energy:
\[U(r) = k\,\frac{(2e)(Ze)}{r},\qquad k = \frac{1}{4\pi\varepsilon_0} = 9\times10^9\]

Step 3: Balance. Energy conservation \(K = U\) gives
\[2eV = k\,\frac{2Ze^2}{r}\]
Dividing both sides by \(2e\),
\[V = k\,\frac{Ze}{r}\;\Rightarrow\; r = \frac{kZe}{V}\]

Step 4: Numerical constant. Insert \(k = 9\times10^9\) and \(e = 1.6\times10^{-19}\) C:
\[r = \frac{(9\times10^9)(1.6\times10^{-19})Z}{V} = \frac{1.44\times10^{-9}\,Z}{V}\ \text{m}\]

Step 5: Angstrom form. Because \(1.44\times10^{-9}\) m equals \(14.4\times10^{-10}\) m and \(10^{-10}\) m \(= 1\) Å,
\[r = \frac{14.4\,Z}{V}\ \text{\AA}\]
Hence proved.
\[\boxed{r = \frac{14.4\,Z}{V}\ \text{\AA}}\]