Question

X and Y are two coplaner and concentric circular loops of radius 10cm and 15cm, having 2 and 3 turns each carrying 5 A current as shown in the figure. Calculate net magnetic field at their common center.

Answer 1

Given:

X and Y are two coplaner and concentric circular loops of radius 10cm and 15cm, having 2 and 3 turns each carrying 5 A current.

To find:

Net magnetic field at the common centre.

Calculation:

Net magnetic field at centre will be sum of individual magnetic field intensity of the coils.

$B = B1 + B2$

$= > B = \dfrac{\mu_{0}}{4\pi} \bigg \{ \dfrac{2\pi i(n1)}{(r1)} \bigg \} + \dfrac{\mu_{0}}{4\pi} \bigg \{ \dfrac{2\pi i(n2)}{(r2)} \bigg \}$

Putting all the available values in SI unit:

$= > B = \dfrac{\mu_{0}}{4\pi} \bigg \{ \dfrac{2\pi i(2)}{(0.10)} \bigg \} + \dfrac{\mu_{0}}{4\pi} \bigg \{ \dfrac{2\pi i(3)}{(0.15)} \bigg \}$

$= > B = \dfrac{\mu_{0}}{4\pi} \bigg \{ \dfrac{2\pi i(2)}{(0.10)} + \dfrac{2\pi i(3)}{(0.15)} \bigg \}$

$= > B = \dfrac{\mu_{0}}{4\pi} \bigg \{ \dfrac{2\pi i}{(0.05)} + \dfrac{2\pi i}{(0.05)} \bigg \}$

$= > B = \dfrac{\mu_{0}}{4\pi} \bigg \{ \dfrac{2\pi i}{(0.05)} \times 2 \bigg \}$

$= > B = \dfrac{\mu_{0}}{4\pi} \bigg \{ \dfrac{4\pi i}{(0.05)} \bigg \}$

$= > B = \dfrac{\mu_{0}}{4\pi} \bigg \{ \dfrac{4\pi i}{5 \times {10}^{ - 2} } \bigg \}$

$= > B = \dfrac{\mu_{0}}{4\pi} \bigg \{ \dfrac{4\pi \times 5}{5 \times {10}^{ - 2} } \bigg \}$

$= > B = \dfrac{\mu_{0}}{4\pi} \bigg \{ \dfrac{4\pi }{ {10}^{ - 2} } \bigg \}$

$= > B = {10}^{ - 7} \times \bigg \{ \dfrac{4\pi }{ {10}^{ - 2} } \bigg \}$

$= > B = 4\pi \times {10}^{ - 5} \: tesla$

So final answer is:

$\boxed{ \sf{B = 4\pi \times {10}^{ - 5} \: tesla}}$