Two circular loops P and Q, each of radius 3 m, are placed coaxially at a separation distance of 4 m. They carry currents of 3 A and 2 A in opposite directions respectively. If the current in loop P is anticlockwise, then the net magnetic field at the centre of the loop P is about
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 5.4 x10-7 T
The field on the axis of a circular loop, at distance x from its centre, is
\( \displaystyle B = \frac{\mu_0 I R^{2}}{2(R^{2} + x^{2})^{3/2}} \)
At the centre of P, loop P gives its centre field (x = 0) and loop Q gives an axial field at x = 4 m. The currents are in opposite directions, so the two fields at the centre of P are opposite and subtract.
\( \displaystyle B_P = \frac{\mu_0 I_P}{2R} = \frac{4\pi\times10^{-7} \times 3}{2 \times 3}\,\mathrm{T} \)
\( \displaystyle B_P = 2\pi\times10^{-7}\,\mathrm{T} \approx 6.28\times10^{-7}\,\mathrm{T} \)
\( \displaystyle B_Q = \frac{4\pi\times10^{-7} \times 2 \times 3^{2}}{2(3^{2} + 4^{2})^{3/2}}\,\mathrm{T} \)
\( \displaystyle B_Q = \frac{4\pi\times10^{-7} \times 18}{2 \times 125}\,\mathrm{T} \approx 0.90\times10^{-7}\,\mathrm{T} \)
\( \displaystyle B_{\mathrm{net}} = B_P - B_Q \)
\( \displaystyle B_{\mathrm{net}} \approx (6.28 - 0.90)\times10^{-7}\,\mathrm{T} \approx 5.4\times10^{-7}\,\mathrm{T} \)
The net field is along the axis in the direction of the field of P (by the right-hand rule, out of the plane of P as viewed when its current appears anticlockwise).