Current density \( \mathbf{j} \) is the current per unit area and is a vector. It is given by:
\[ \mathbf{j} = \frac{I}{A} \]
The drift velocity \( v_d \) is related to the electric field \( E \) as:
\[ v_d = -\frac{eE\tau}{m} \]
Where \( e \) is the charge of an electron, \( \tau \) is the relaxation time, and \( m \) is the mass of the electron.
The current density is given by the product of the number of charge carriers \( n \), the charge of the carriers \( e \), and their drift velocity \( v_d \):
\[ \mathbf{j} = n(-e)v_d = n \cdot (-e) \cdot \left( -\frac{eE\tau}{m} \right) \]
Simplifying this, we get:
\[ \mathbf{j} = \frac{ne^2\tau E}{m} \]
Thus, the conductivity \( \alpha \) is:
\[ \alpha = \frac{ne^2\tau}{m} \]
The Wheatstone bridge is used to measure an unknown resistance. It is balanced when the following condition is met:
\[ \frac{R_1}{R_2} = \frac{R_3}{R_4} \]
In this case, there is no current flowing through the galvanometer, and the bridge is considered balanced.
An infinitely long straight wire carrying current $I$ is bent in a planar shape as shown in the diagram. The radius of the circular part is $r$. The magnetic field at the centre $O$ of the circular loop is :

A racing track is built around an elliptical ground whose equation is given by \[ 9x^2 + 16y^2 = 144 \] The width of the track is \(3\) m as shown. Based on the given information answer the following: 
(i) Express \(y\) as a function of \(x\) from the given equation of ellipse.
(ii) Integrate the function obtained in (i) with respect to \(x\).
(iii)(a) Find the area of the region enclosed within the elliptical ground excluding the track using integration.
OR
(iii)(b) Write the coordinates of the points \(P\) and \(Q\) where the outer edge of the track cuts \(x\)-axis and \(y\)-axis in first quadrant and find the area of triangle formed by points \(P,O,Q\).