Consider a circular coil of radius 'r' and carrying current 'I' as shown in fig.

The magnet flux density at the center of the coil is given as -

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Question

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RSMSSB JE Electrical/Mechanical 2020 Official Paper (Degree): Held on 26 Dec 2020

Answer 1

Concept:

Magnetic flux density for any current-carrying element is given by:

B = \(( {μ_o\over 4π } \times {I\over r })\times ϕ \)

where, B = Magnetic flux density

μ_o = absolute permeability

I = current flowing in current-carrying element

r = distance from current-carrying element to required point

ϕ = angle made by current-carrying element with respect to the required point

Explanation:

F1 Jai P 28-2-22 Savita D3

B_net = B₁ + B₂ + B₃

Magnetic flux densities B₁ and B₃ are equal to zero as the current flowing is equal and opposite in nature.

So, B_netwill be due to circular loop only.

Hence, ϕ = 2π

Bnet = B₂

B_net= \(( {μ_o\over 4π } \times {I\over r })\times 2\pi \)

Bnet = \( {μ_o\over 2 } {I\over r }\)

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