Two concentric circular coils $X$ and $Y$ of radii $16\; cm$ and $10\;cm$ respectively, lie in the same vertical plane containing the north to south direction. Coil $X$ has $20$ turns and carries a current of $16\; A$ coil $Y$ has $25$ turns and carries a current of $18\; A$. The sense of the current in $X$ is anticlockwise, and clockwise in $Y$, for an observer looking at the coils facing west. Give the magnitude and direction of the net magnetic field due to the coils at their centre.
Two concentric circular coils $X$ and $Y$ of radii $16\; cm$ and $10\;cm$ respectively, lie in the same vertical plane containing the north to south direction. Coil $X$ has $20$ turns and carries a current of $16\; A$ coil $Y$ has $25$ turns and carries a current of $18\; A$. The sense of the current in $X$ is anticlockwise, and clockwise in $Y$, for an observer looking at the coils facing west. Give the magnitude and direction of the net magnetic field due to the coils at their centre.
Radius of coil $X , I _{1}=16 \,cm =0.16 \,m$
Radius of coil $Y , r_{2}=10 \,cm =0.1 \,m$
Number of turns of on $\operatorname{coil} X , n_{1}=20$
Number of turns of on coil $Y , n_{2}=25$
Current in $\operatorname{coil} X , I_{1}=16 \,A$
Current in coil $Y , I_{2}=18 \,A$
Magnetic field due to coil $X$ at their centre is given by the relation,
$B_{1}=\frac{\mu_{0} n_{1} I_{1}}{2 r_{1}}$
Where, $\mu_{0}=$ Permeability of free space $=4 \pi \times 10^{-4} \,T\,m\, A ^{-1}$
$\therefore B_{1}=\frac{4 \pi \times 10^{-7} \times 20 \times 16}{2 \times 0.16}$
$=4 \pi \times 10^{-4} \,T$ (towards East) Magnetic field due to coil Y at their centre is given by the relation,
$B_{2}=\frac{\mu_{0} n_{2} I_{2}}{2 r_{2}}$
$=\frac{4 \pi \times 10^{-7} \times 25 \times 18}{2 \times 0.10}$
$=9 \pi \times 10^{-4} \,T$ (towards West)
Hence, net magnetic field can be obtained as:
$B=B_{2}-B_{1}$
$=9 \pi \times 10^{-4}-4 \pi \times 10^{-4}$
$=5 \pi \times 10^{-4}\, T$
$=1.57 \times 10^{-3}\; T$ (towards West)
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