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: Statement of Biot-Savart law. Consider a small element \( d\vec{l} \) of a conductor carrying current \( I \). The tiny magnetic field \( dB \) produced by this element at a point \( P \), whose position vector from the element is \( \vec{r} \) (making angle \( \theta \) with \( d\vec{l} \)), is:
(a) directly proportional to the current \( I \);
(b) directly proportional to the length \( dl \) of the element;
(c) directly proportional to \( \sin\theta \);
(d) inversely proportional to the square of the distance \( r \).
\[ dB = \frac{\mu_0}{4\pi}\,\frac{I\,dl\,\sin\theta}{r^2} \]
In vector form \( d\vec{B} = \dfrac{\mu_0}{4\pi}\dfrac{I\,d\vec{l}\times\hat{r}}{r^2} \). The direction of \( dB \) is perpendicular to the plane containing \( d\vec{l} \) and \( \vec{r} \) (right-hand rule). Here \( \mu_0 = 4\pi\times10^{-7}\,\text{T m A}^{-1} \) is the permeability of free space.

Step 2: Set-up for a straight wire. Let a long straight conductor carry current \( I \). Let \( P \) be a point at perpendicular distance \( a \) from the wire, with foot of perpendicular \( O \). Consider an element \( d\vec{l} \) at distance \( l \) from \( O \). Let \( \phi \) be the angle that the line joining the element to \( P \) makes with the perpendicular \( OP \).

Step 3: Geometry. From the figure, \( l = a\tan\phi \), so \( dl = a\sec^2\phi\,d\phi \); the distance \( r = a\sec\phi \); and the angle between \( d\vec{l} \) and \( \vec{r} \) is \( \theta = 90^\circ - \phi \), so \( \sin\theta = \cos\phi \).

Step 4: Field of the element. Substituting into Biot-Savart law:
\[ dB = \frac{\mu_0}{4\pi}\,\frac{I\,(a\sec^2\phi\,d\phi)(\cos\phi)}{(a\sec\phi)^2} = \frac{\mu_0 I}{4\pi a}\cos\phi\,d\phi \]

Step 5: Integrate over the whole wire. If the wire subtends angles \( \phi_1 \) and \( \phi_2 \) on the two sides of \( O \):
\[ B = \frac{\mu_0 I}{4\pi a}\int_{-\phi_1}^{\phi_2}\cos\phi\,d\phi = \frac{\mu_0 I}{4\pi a}\,(\sin\phi_1 + \sin\phi_2) \]

Step 6: Infinitely long wire. For a very long (infinite) wire, \( \phi_1 = \phi_2 = 90^\circ \), so \( \sin\phi_1 + \sin\phi_2 = 2 \):
\[ B = \frac{\mu_0 I}{4\pi a}\times 2 = \frac{\mu_0 I}{2\pi a} \]
The field lines are concentric circles around the wire.
\[\boxed{B = \frac{\mu_0 I}{2\pi a}}\]

OPTION 2: Distance of Closest Approach of α-particle

Step 1: Charge and energy of the α-particle. An α-particle carries charge \( q = 2e \). When it is accelerated through a potential difference \( V \) volt, the kinetic energy gained is
\[ K = qV = 2eV \quad \text{(in joule)} \]

Step 2: Condition at closest approach. As the α-particle moves toward the nucleus (charge \( +Ze \)), the Coulomb repulsion slows it down. At the nearest distance \( r \) it momentarily stops, so all its kinetic energy has become electrostatic potential energy:
\[ U = \frac{1}{4\pi\varepsilon_0}\,\frac{(2e)(Ze)}{r} = \frac{1}{4\pi\varepsilon_0}\,\frac{2Ze^2}{r} \]

Step 3: Equate energy. Setting \( K = U \):
\[ 2eV = \frac{1}{4\pi\varepsilon_0}\,\frac{2Ze^2}{r} \]
Cancel \( 2e \) from both sides:
\[ V = \frac{1}{4\pi\varepsilon_0}\,\frac{Ze}{r} \quad\Rightarrow\quad r = \frac{1}{4\pi\varepsilon_0}\,\frac{Ze}{V} \]

Step 4: Put in the numbers. Using \( \dfrac{1}{4\pi\varepsilon_0} = 9\times10^{9}\ \text{N m}^2\text{C}^{-2} \) and \( e = 1.6\times10^{-19}\ \text{C} \):
\[ r = \frac{9\times10^{9}\times(1.6\times10^{-19})\,Z}{V} = \frac{1.44\times10^{-9}\,Z}{V}\ \text{m} \]

Step 5: Convert to angstrom. Since \( 1\ \text{\AA} = 10^{-10}\ \text{m} \), \( 1.44\times10^{-9}\ \text{m} = 14.4\times10^{-10}\ \text{m} = 14.4\ \text{\AA} \). Therefore
\[ r = \frac{14.4\,Z}{V}\ \text{\AA} \]
which is the required result.
\[\boxed{r = \frac{14.4\,Z}{V}\ \text{\AA}}\]