# 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

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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.

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

B 2πr = μJ ⋅ πr2

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