Question : Consider the circuit shown where APB and AQB are semicircles. What will be the magnetic field at the centre C of the circular loop?
Doubt by Tanisha
Text Solution :
We know, Magnetic field due to a circular current carrying loop at the centre is given by μ₀I/2a where a is the radius of the circular current carrying coil.
Now,
Magnetic field due to semicircular wire APB
BAPB = μ₀I/4a (Inside the plane of the screen)
Similarly
BAQB = μ₀I/4a (Outside the plane of screen)
We know, Magnetic field due to a circular current carrying loop at the centre is given by μ₀I/2a where a is the radius of the circular current carrying coil.
Now,
Magnetic field due to semicircular wire APB
BAPB = μ₀I/4a (Inside the plane of the screen)
Similarly
BAQB = μ₀I/4a (Outside the plane of screen)
Both magnetic fields have equal magnitudes but they are opposite in direction, that's why they will cancel out each other. Hence, net magnetic field at the centre will be zero.
Video Solution :
