Question

Two concentric circular coils of ten turns each are situated in the same plane. Their radii are $20$ and $40\,cm$ and they carry respectively $0.2$ and $0.4$ ampere current in opposite directions. The magnetic induction (in tesla) at the centre is:
A. ${\text{ }}\dfrac{{3{\mu _0}}}{4}$
B. ${\text{ }}\dfrac{{5{\mu _0}}}{4}$
C. ${\text{ }}\dfrac{{7{\mu _0}}}{4}$
D. ${\text{ }}\dfrac{{9{\mu _0}}}{4}$

Answer 1

in order to solve the question, we will first find the magnetic induction in the first coil after then we will find the magnetic induction in the second coil as both the coils are concentric we have to subtract the magnetic induction of first coil from the second coil and we will find the net magnetic induction.

Formula used:
Magnetic induction at the centre of a current carrying loop
$B = \dfrac{{{\mu _0}NI}}{{2R}}$
Where, $B$ represents the magnetic induction,$N$ represents the number of turns in the coil, $I$ represents the current of the coil, $R$ represents the radius of the coil and ${\mu _0}$ represents the permeability constant.

Complete step by step answer:
In the question we are given that there are two concentric circular coils of ten turns each are situated in the same plane radius of the both coils and along with that the current in the both coil is also given and we have to find the magnetic induction and the centre of both concentric coils
Radius of coil 1 = $20\,cm$
Radius of coil 2= $40\,cm$
Current in the coil 1 = $0.2\,A$
Current in the coil 2 = $0.3\,A$
At the centre of a current carrying loop magnetic induction is given by
$B = \dfrac{{{\mu _0}NI}}{{2R}}$

Coil 1:
Radius: ${R_1}{\text{ }} = {\text{ }}20{\text{ }}cm$ = 0.2 m
Current: ${I_1}{\text{ }} = {\text{ 0}}.2A$
Number of turns in the coil: N = 10
Using the formula for coil 1
${B_1} = \dfrac{{{\mu _0}{N_1}{I_1}}}{{2R}}$
Substituting the values in the above equation we get
${B_1} = \dfrac{{{\mu _0} \times 10 \times 0.2}}{{2 \times 0.2}}$
Hence the value of magnetic induction in coil 1
${B_1} = 5{\mu _0}$T

Coil 2:
Radius: ${R_1}{\text{ }} = {\text{ 4}}0{\text{ }}cm$ = 0.4 m
Current: ${I_1}{\text{ }} = {\text{ 0}}{\text{.3}}A$
Number of turns in the coil: N = 10
Using the formula for coil 2
${B_2} = \dfrac{{{\mu _0}{N_2}{I_2}}}{{2R}}$
Substituting the values in the above equation we get
${B_2} = \dfrac{{{\mu _0} \times 10 \times 0.3}}{{2 \times 0.4}}$
Hence the value of magnetic induction in coil 1
${B_2} = \dfrac{{15{\mu _0}}}{4}$T

Net magnetic induction at the centre of concentric circular coils
${B_{NET}} = {B_1} - {B_2}$
Substituting the values of magnetic induction of coil 1 and coil 2
${B_{NET}} = 5{\mu _0} - \dfrac{{5{\mu _0}}}{4}$
${B_{NET}} = \dfrac{{5{\mu _0}}}{4}$T
Net magnetic induction at the centre of concentric circular coils ${B_{NET}} = \dfrac{{5{\mu _0}}}{4}$T.

Hence, the correct option is B.

Note: Many of the students will make the mistake instead of subtracting the magnetic induction both the coils they can add the magnetic induction of the coils but as given in the question coils are in the opposite direction so we have to subtract them but if the coils are in the same direction we have to add them.