Electric current is passing through a solid conductor PQ from P to Q. The electric
current densities at P and Q are in the ratio:

A. 1 : 2
B. 2 : 1
C. 1 : 4
D. 4 : 1

Hint:The current density is the ratio of current to the cross sectional area. Express the current density at point P and Q. Take the ratio of these current densities to get the required answer. The area of the circular loop of radius R is \[\pi {R^2}\].

Formula used:
Current density, \[j = \dfrac{I}{A}\]

Here, I is the current and A is the area.

Complete step by step answer:

We know when the current flows through a given area, the current density in this area is
expressed as,
\[j = \dfrac{I}{A}\]

Here, I is the current and A is the area.

We can see in the figure, the area of the circular loop at position P and Q is different. Let’s
express the current density at position P as follows,
\[{j_1} = \dfrac{I}{{\pi {{\left( {2R} \right)}^2}}}\]
\[ \Rightarrow {j_1} = \dfrac{I}{{4\pi {R^2}}}\] …… (1)

Let’s express the current density at position Q as follows,
\[{j_2} = \dfrac{I}{{\pi {R^2}}}\] …… (2)

Dividing equation (1) by equation (2), we get,
\[\dfrac{{{j_1}}}{{{j_2}}} = \dfrac{{\dfrac{I}{{4\pi {R^2}}}}}{{\dfrac{I}{{\pi {R^2}}}}}\]
\[ \Rightarrow \dfrac{{{j_1}}}{{{j_2}}} = \dfrac{1}{4}\]
\[ \therefore {j_1}:{j_2} = 1:4\]

So, the correct answer is option (C).

Note:Students often don’t get the difference between the current and current density. The current is the rate of flow of charges per unit time. The current density is the current flowing per unit area of the cross section. If the area of the cross section is greater, the current density decreases. Remember, the current density is also a vector quantity like current.