JEE Main · 2023 · 31 Jan 2023 · Shift 1 · Magnetic Properties of Matter

A rod with circular cross-section area

2

^{2}

and length 40 cm is wound uniformly with 400 turns of an insulated wire. If a current of 0.4 A flows in the wire windings, the total magnetic flux produced inside windings is

4 π \times 1 0^{- 6}

Wb. The relative permeability of the rod is
(Given : Permeability of vacuum

μ_{0} = 4 π \times 1 0^{- 7}

^{- 2}

)

(A)12.5
(B) $\frac{32}{5}$
(C)125✓ Correct
(D) $\frac{5}{16}$

Answer: (C)

Solution

The flux linked with such a winding is

ϕ = μ_{r} μ_{0} \frac{N}{ℓ} I A

. Solving for the relative permeability,

μ_{r} = \frac{ϕ}{μ _{0} \frac{N}{ℓ} I A}

. With

ϕ = 4 π \times 1 0^{- 6}

Wb,

μ_{0} = 4 π \times 1 0^{- 7}

N = 400

ℓ = 0.4

I = 0.4

A and

A = 2 \times 1 0^{- 4}

^{2}

, the denominator is

4 π \times 1 0^{- 7} \times 1000 \times 0.4 \times 2 \times 1 0^{- 4} = 32 π \times 1 0^{- 9}

, so

μ_{r} = \frac{4 π \times 1 0 ^{- 6}}{32 π \times 1 0 ^{- 9}} = 125

JEE Main · 2023 · 31 Jan 2023 · Shift 1 · Single correct option

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JEE Main · 2023 · 31 Jan 2023 · Shift 1 · Magnetic Properties of Matter

Solution

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