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PGCIL DT Electrical 13 Aug 2021 Official Paper (NR I)

Option 4 : 0.63 m

**Concept:**

__1. Magnetic Field Intensity or Magnetizing force due to long straight line conductor By using Ampere Circuital Law:__

Ampere’s circuital law states that line integral of the magnetic field forming a closed loop around the current (I) carrying wire, in the plane normal to the current, is equal to the μo times the net current passing through the close loop.

\(\mathop \oint \limits_C^{} B.dl = {μ _o}i\)

For Infinite Straight Line Conductor:

\(\mathop \oint \limits_C^{} B.dl =Bl=2π RB\)

Where, l = 2πR

Hence, B = \(\frac{μ_oi}{2π R}\)

We know that, B = μ_{0}H

∴ \(H=H_s=\frac{i}{2π R}\)

__2. Magnetic Field at the center of a Circular Current carrying coil:__

Consider a current-carrying circular loop having its center at O carrying current i as shown below,

If dl is a small element at a distance r then, the magnetic field intensity on that point can be written using Biot-Savart’s law.

\(|\vec{dB}|=\frac{\mu_0i}{4π}\frac{idlsinθ}{r^2}\) .... (1)

Since, the loop is circular,

θ = 90° → sin θ = 1

Hence, equation (1) becomes,

\(|\vec{dB}|=\frac{\mu_0i}{4π}\frac{idl}{r^2}\)

If the circular loop is consisting of numbers of small element dl, then we will get the magnetic intensity all over the loop.

Hence to get the total field we must sum up that is integrate the magnetic field all over the field.

\(B=∫{\vec{dB}}\)

\(B=∫|\vec{dB}|=∫\frac{\mu_0i}{4π}\frac{idl}{r^2}\)

\(B=\frac{\mu_0i}{4π r^2}∫ dl\)

∫dl = l = 2πr

Hence,

\(B=\frac{\mu_oi}{2r}\)

Now, we have to find magnetic field intensity or magnetizing force (H),

And, **H = H _{C} = **\(\frac{B}{\mu_0}=\frac{i}{2r}\)

**Application:**

Let H_{C} is the magnetizing force due to circular loop

And H_{s} is the magnetizing force due to the straight-line conductor.

According to the question,

H_{C} = H_{s} .... (1)

Let R is the length of point from point where force produced by the straight conductor = 0.1 m

r is the radius of circular loop = ?

i = 10 A

From equation (1),

\(\large{\frac{i}{2r}=\frac{i}{2π R}}\)

or 2r = 2πR

r = πR

Since, R = 0.1 m

∴ r = π × 0.1 = 0.314 m

We know that, Diameter (D) = 2 × Radious (r)

**Hence, Diameter (D) = 2 × 0.314 ≈ 0.63 m**