Fig. 28-9 shows the cross-section of a hollow cylinder of inner radius a = 5 cm and outer radius b = 7 cm. a uniform current density j = 1 a/cm2 flows through the cylinder. calculate the magnitude of the magnetic field at a distance d = 10 cm from the axis of the cylinder.

1.507×10⁻⁴ T

Explanation:

a = 5 cm = 0.05 m

b = 7 cm = 0.07 m

j = 1 A/cm²

Distance from magnetic field = d = 10 cm = 0.1 m

μ₀ = Vacuum permeability = 4π×10⁻⁷ H/m

Magnetic of hollow cylinder

∴ Magnitude of the magnetic field at a distance d = 10 cm from the axis of the cylinder is 1.507×10⁻⁴ T

the rearrangement of particles in a physical change. hope this

speed of light in vacuum is 300,000 km/s

speed of light in other mediums is given by the expression v = c/n, where c is the speed of light in vacuum and n is the refractive index of medium.

for the first medium v = 200,000 km/s

  so, refractive index of medium = 300000/200000 = 1.5

    1.5 refractive index is usually for glass. so first medium is glass.

for the second medium v = 250,00 km/s

so, refractive index of medium = 300000/250000 = 1.2

    1.2 refractive index is usually for water. so second medium is water.

for the third medium v = 0 km/s

so, refractive index of medium = 300000/0 = ∞

    ∞ refractive index is usually for opaque mediums. so third medium is opaque medium.