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A current of 5 A passes along the axis of a cylinder of 5 cm radius. The flux density at the surface of the cylinder is
1. 2 μT
2. 20 μT
3. 200 μT
4. 2000 μT

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Correct Answer - Option 2 : 20 μT

Concept:

The magnetic flux density inside a cylinder with uniform current density is given by:

\(B = \frac{{{\mu _0}rI}}{{2\pi {R^2}}}\)

r = distance from the center of the cylinder (r < R)

R = Radius of the cylinder

I = Current flowing across the cylinder

Application:

At the surface of the cylinder, r = R, and the flux density becomes:

\(B = \frac{{{\mu _0}RI}}{{2\pi {R^2}}} = \frac{{{\mu _0}I}}{{2\pi {R}}}\)

With I = 5 A, and R = 5 × 10-2 m, the flux density at the surface will be:

\(B = \frac{{{4\pi × 10^{-7}× 5}}}{{2\pi ×{5× 10^{-2}}}}\)

B = 20 × 10-6 T

B = 20 μT

Important Derivation:

Ampere's Circuital law states that the line integral of the magnetic field of induction \(\vec B\) around any closed path in free space is equal to absolute permeability of free space μ0 times the total current flowing through the area bounded by the path.

Mathematically:

\(\mathop \oint \nolimits \vec B \cdot d\vec l = {\mu _0}I\)

\(\mathop \oint \nolimits \vec B \cdot d\vec l = {\mu _0}\mathop \smallint \nolimits \vec J \cdot d\vec s\)

For uniform current distribution I, the current density will be:

\(J = \frac{I}{{\pi {R^2}}}\)

The current at a distance r will be:

\(I = \frac{I\pi r^2}{{\pi {R^2}}}=\frac{Ir^2}{{ {R^2}}}\)

The current enclosed increases as the square of the radial distance.

R = radius of cylinder

Now, Applying Amperes circuital law, the magnetic field at a distance r inside a conductor of radius R will be:

B 2πr = μJ ⋅ πr2

\(B = \frac{{{\mu _0}rI}}{{2\pi {R^2}}}\)

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