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  • 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

4.14) 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.

Answers (1)

best_answer

Using the right-hand thumb rule the direction of the magnetic field due to coil X and Y is shown in the figure below as Bx and By.

So, the magnetic field due to coil X at the centre is given by

$$
\begin{aligned}
B_x & =\frac{\mu_0 n_x i_1}{2 r_x} \\
B_x & =\frac{\left(4 \pi \times 10^{-7}\right) \times 20 \times 16}{2 \times\left(16 \times 10^{-2}\right)} \\
B_x & =\left(4 \pi \times 10^{-4}\right)(\text { Towards East })
\end{aligned}
$$


Similarly, the magnetic field at the common centre due to coil Y ,

$$
\begin{aligned}
& B_y=\frac{\mu_0 n_y i_2}{2 r_y n j} \\
& B_y=\frac{\left(4 \pi \times 10^{-7}\right) \times 25 \times 18}{2 \times\left(10 \times 10^{-2}\right)} \\
& B_y=9 \pi \times 10^{-4} \mathrm{~T} \\
& \text { (TowardsWest) }
\end{aligned}
$$


So, the net magnetic field at the common centre is

$$
\begin{aligned}
& B_{\text {net }}=B_y-B_x \quad\left(\text { as } B_y>B_x\right) \\
& B_{\text {net }}=\left(9 \pi \times 10^{-4}\right)-\left(4 \pi \times 10^{-4}\right) \\
& B_{\text {net }}=5 \pi \times 10^{-4} \mathrm{~T} \text { (TowardsWest) }
\end{aligned}
$$


Hence, the magnitude of the net magnetic field is due to the coils at their centre towards the west, is $B_{\text {net }}=5 \pi \times 10^{-4} \mathrm{~T}$.

 

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Sayak

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