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For a given ferrite material operating in linear mode, B = 0.05 T and μr = 50. The magnetic field H is
mA/m
125 A/m
796 A/m
1000 A/m
796 A/m
Given: Magnetic flux density B = 0.05 T, relative permeability μr = 50, and permeability of free space μ0 = 4π × 10°-7 H/m.
Magnetic flux density B = 0.05 T, relative permeability μr = 50, and permeability of free space μ0 = 4π × 10°-7 H/m.
H=μ0×μrB
Identify the relationship between B and H
The magnetic field intensity H is related to the magnetic flux density B by the permeability of the material μ, where μ=μ0μr.
B=μH=μ0μrH
Rearrange the formula to solve for H
Isolate H by dividing both sides by the permeability of the medium μ0μr.
H=μ0μrB
Substitute the given values
Plug in B=0.05 T, μ0=4π×10°−7 H/m, and μr=50.
H=4π×10°−7×500.05
Calculate the result
Simplify the expression: H=200π×10°−70.05=0.000062830.05≈795.77 A/m. Rounding to the nearest integer gives 796 A/m.
H≈796 A/m
C is correct because calculating the magnetic field intensity using the formula H=μ0μrB yields approximately 796 A/m.
This formula is fundamental in transformer and motor design; remember that μ0 is constant while μr depends on the core material saturation characteristics.