Question

A rod with circular cross-section area $2\, cm ^2$ 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 $4 \pi \times 10^{-6}\, Wb$ The relative permeability of the rod is(Given : Permeability of vacuum $\mu_0=4 \pi \times 10^{-7}\, NA ^{-2}$ )

• A. $125$
• B. $12.5$
• C. $\frac{5}{16}$
• D. $\frac{32}{5}$

Hint:

The magnetic flux inside a coil is proportional to the current, the number of turns, the cross-sectional area, and the relative permeability of the material.

Answer 1

The Correct Option is A

ϕ = μ_r​μ₀​\(\frac nl\)​I × A 

ϕ = 4π × 10−⁶ x 4π × 10⁻⁷ \(\frac {400}{0.40}\) x 0.4 x 2 x 10 ^-4      
μ_r​=125

So, the correct answer is (A): 125

Answer 2

The correct option is (A): 125

By applying the Magnetic flux concept,
\(Φ = B.ACos \theta\)

As the area vector and magnetic field are tangential, both will be in a horizontal direction,
\(Φ = B.ACos0\degree\)
\(Φ = B.A\), where A is the cross-section area and 
\(B = μ. n. i\), where again \(μ = μ_r. μ_0\)
μ = permeability of the
μ_r = relative permeability,
μ₀ = permeability of vacuum,
n = no. of times,
i = current,

A = Area
\(Φ = μ_r . μ_0 . n . i . A\)
\(Φ=4\pi \times 10^{-6} \times 4\pi \times 10^{-7} \times \frac{400}{0.40} \times 0.4 \times 2 \times 10^{-4}\)
\(\rightarrow μ_r = \frac{100}{0.8}\)
\(\rightarrow μ_r = 125\)

Therefore, the Relative Permeability of the rod = 125