Solution (a)
Step 1 (Definition and Nature): [0.5 Mark]
Current density ($\vec{j}$) is defined as the amount of electric current flowing per unit area of cross-section held normal to the direction of the current. It is a vector quantity.
Mathematically: $\vec{j} = \frac{I}{A} \hat{n}$
Step 2 (Mathematical Derivation): [2 Marks]
- The drift velocity ($\vec{v}_d$) of electrons in an electric field $\vec{E}$ is given by: $$\vec{v}_d = -\frac{e\vec{E}}{m}\tau$$
- The relation between current density ($\vec{j}$) and drift velocity ($\vec{v}_d$) is: $$\vec{j} = -ne\vec{v}_d$$
- Substituting the expression for $\vec{v}_d$ into the equation for $\vec{j}$: $$\vec{j} = -ne \left( -\frac{e\vec{E}}{m}\tau \right)$$
- Simplifying the expression: $$\vec{j} = \left( \frac{ne^2\tau}{m} \right) \vec{E}$$
Step 3 (Final Conclusion): [0.5 Mark]
Comparing the derived expression with $\vec{j} = \alpha \vec{E}$, we get:
$$\alpha = \frac{ne^2\tau}{m}$$This constant $\alpha$ is known as the electrical conductivity ($\sigma$) of the material.
💡 EL10 Concept Insight
Think of current density as the "crowd flow" of electrons: just like more people passing through a gate per second means a higher flow, the electric field acts like a push that accelerates electrons, while the relaxation time is the average time they can "run" before bumping into atoms and slowing down.
Solution (b)
Step 1 (Definition): [0.5 Mark]
A Wheatstone bridge is an electrical circuit used to measure an unknown resistance by balancing two legs of a bridge circuit, one leg of which includes the unknown component.
Step 2 (Mathematical Working - Kirchhoff's Laws): [2 Marks]
- Consider a bridge with resistors $P, Q, R, S$ and a galvanometer $G$. Applying Kirchhoff’s Junction Rule at nodes B and D:
At node B: $I_1 - I_g - I_3 = 0$
At node D: $I_2 + I_g - I_4 = 0$ - Applying Kirchhoff’s Loop Rule to loop ABDA: $$-I_1P - I_gG + I_2R = 0$$
- For a balanced bridge, the galvanometer shows no deflection, so $I_g = 0$. The equations become:
$I_1 = I_3$ and $I_2 = I_4$
$-I_1P + I_2R = 0 \implies I_1P = I_2R$ - Applying Kirchhoff’s Loop Rule to loop BCDB: $$-I_3Q + I_4S + I_gG = 0$$ Since $I_g = 0$, then $I_3Q = I_4S$.