Question

There are two separate long cylindrical wires having uniform current density. The radius of one of the wires is twice that of the other. The figure shows the plot of magnitude of magnetic field intensity versus radial distance $\left(r\right)$ from their axis. If the curved parts of the two graphs are overlapping , then which of the following options is/are correct

$[$ Note $:-j$ is the current density which is equal to $\frac{I}{A}]$

• A. $j_1>j_2$
• B. $R_2=2R_1$
• C. $I_1=I_2$
• D. $B_1=2B_2$

Answer 1

Answer: (A), (B), (C), (D)

$B_1=2B_2$

$\text{STEP-I}:-$

Field at a point $P_1(x<R)$ is $$B=\frac{{\mu }_{o}jx}{2}$$$\Rightarrow B\propto x$

So, we can say that the slope of $1^{st}$ graph is higher , $\therefore j_1>j_2$

Option (a) is correct

$\text{STEP-II}:-$

From the graph as well as from the data given in the question we can conclude that , $R_2=2R_1$

Option (b) is correct.

$\text{STEP-III}:-$

Field at point $P_2(x>R)$ is,$$B=\frac{{\mu }_{0}I}{2\pi x}$$Since the two graphs overlap, we can say that $I_1=I_2$

Option (c) is correct.

$\text{STEP-IV}:-$

Field on the surfaces of the two wires are :- $B_1=\frac{{\mu }_0{I}_1}{2\pi R_1}$ and $B_2=\frac{{\mu }_0{I}_2}{2\pi R_2}$

$\therefore \frac{B_1}{B_2}=\frac{R_2}{R_1}=\frac{2}{1}$

$\Rightarrow B_1=2B_2$

Option (d) is correct.

Hence, all options are correct alternatives for the given question.