Question

The given diagram is a cross-sectional view of coaxial cable. The central conductor is surrounded by a rubber layer, which is surrounded by an outer conductor, which is further surrounded by another rubber layer. Current in the inner conductor is $1\text{A}$ directed out of the page, and current in the outer conductor is $3\text{A}$ into the page. The ratio of magnitude of magnetic field at point $\text{a}$ to that at point $\text{b}$ will be -

• A. $1$
• B. $2$
• C. $1.5$
• D. $2.5$

Answer 1

The correct option is C $1.5$


According to Ampere's circuital law, $$\oint \overrightarrow{B}\cdot \overrightarrow{dl}={\mu }_{o}I$$
Consider a circular closed path of radius $1\text{mm}$, oriented anticlockwise, such that point $a$ lies on its circumference.

Then, $\oint {\overrightarrow{B}}_a\cdot \overrightarrow{dl}=\mu I_1$

$\Rightarrow B_a(2\pi \times {10}^{-3})={\mu }_0{I}_1$

$\Rightarrow B_a(2\pi \times {10}^{-3})=4\pi \times {10}^{-7}\times 1$

$\Rightarrow B_a=2\times {10}^{-4}\text{T}$

Similarly, consider a circular closed path of radius $3\text{mm}$, oriented clockwise, such that point $b$ lies on its circumference.

Then, $\oint {\overrightarrow{B}}_b\cdot \overrightarrow{dl}={\mu }_o(I_2-I_1)$

$\Rightarrow B_b(2\pi \times 3\times {10}^{-3})={\mu }_o(I_2-I_1)$

$\Rightarrow B_b(2\pi \times 3\times {10}^{-3})=4\pi \times {10}^{-7}\times (3-1)$

$\Rightarrow B_b=\frac{4}{3}\times {10}^{-4}\text{T}$

Hence, the ratio of magnitude of the magnetic field at point $a$ to that at point $b$ will be,

$\frac{B_a}{B_b}=\frac{2\times {10}^{-4}}{4/3\times {10}^{-4}}=1.5$

Hence, option $\left(C\right)$ is the correct answer.