Question

. Two circular loops of radii 6.28 cm and 3.14 cm are arranged concentric.to one another
with their planes at right angles to each other. If a current of 2 A is passed through each
of them, calculate the magnitude of the magnetic field at their common centre
(Given: Ho = 411 x 10-7 Hm-1).

Answer 1

Explanation:

i hole it's help you

pls make me as Briliantist

Answer 2

Given:

Radius of circular loop 1 = 6.28cm

Radius of circular loop 2 = 3.14cm

The current passing through each of them = 2A

Value of μ$_{o}$ = 4π × $10^{-7}$ Wb/ A.m

To Find:

The value of the magnetic field at their common center.

Solution:

The value of the magnetic field due to the circular coil,

B = μ$_{o}$ × $\frac{i}{2r}$

The value of the magnetic field of loop 1 (B$_{1}$) = μ$_{o}$×$\frac{2}{2*6.28}$ = 2×$10^{-7}$ Wb/m²

Similarly, The value of B$_{2}$ = 4×$10^{-7}$ Wb/m²

Both the magnetic fields are right angles to each other.

So, the net magnetic field (B) = √(B1²+ B2²) = 2√5 × $10^{-7}$ Wb/m²

Hence. the magnitude of the net magnetic field at their common center is 2√5×$10^{-7}$ Wb/m².