Question

Two circular coils of radii 5.0 cm and 10 cm carry equal currents of 2.0 A. The coils have 50 and 100 turns respectively and are placed in such a way that their planes as well as the centres coincide. If the outer coil is rotated through 90° about a diameter, Find the  magnitude of the magnetic field B at the common centre of the coils if the currents in the coils are (a) in the same sense (b) in the opposite sense.

Answer 1

Given:-

No. of turns: n₁ = 50 and n₂ = 100

Magnitude of currents: i₁ = i₂ = 2 A

Radii of loops: r₁ = 5 cm and r₂ = 10 cm

(a) In the same sense:-

The magnetic field intensity at the centre due to C₁ is given by

\[B_1 = \frac{\mu_0 n_1 i_1}{2 r_1}\]

\[ = \frac{4\pi \times {10}^{- 7} \times 50 \times 2}{2 \times 5 \times {10}^{- 2}}\]

\[ = 4\pi \times {10}^{- 4} \] T

(In the plane of paper in upward direction)

The magnetic field intensity at the centre due to C₂ is given by

\[B_2 = \frac{\mu_0 n_2 i_2}{2 r_2}\]

\[ = \frac{4\pi \times {10}^{- 7} \times 100 \times 2}{2 \times 10 \times {10}^{- 2}}\]

\[ = 4\pi \times {10}^{- 4}\] T

(In the plane of paper in upward direction)

In this case, magnetic fields due to C₁ and C₂ at the centre are along the same direction.

Thus, the net magnetic field is given by

\[B_{net} = B_1 + B_2 \]

\[ = (4\pi \times {10}^{- 4} ) + (4\pi \times {10}^{- 4} )\]

\[ = 8\pi \times {10}^{- 4} \] T

\[ = 25 . 12 \] mT

(b) When the direction of current in the two coils is opposite to each other then the magnetic fields will also point in opposite directions as shown in the figure. Hence, the net magnetic field will be obtained by the subtraction of the two magnetic fields.

\[B_{net} = B_1 - B_2 \]

\[ = (4\pi \times {10}^{- 4} ) - (4\pi \times {10}^{- 4} )= 0\]