Question 7.10: The energy stored per unit volume of magnetic field in air i...

W_{ƒ}  (air) = \frac{B²}{2 µ_{0}} J/m³

For cross-section 1 m² and length l of air distance

W_{ƒ}  (B, l) = \left(\frac{B²}{2 µ_{0} }\right)  l,

Force of attraction

F =  \frac{\partial W_{ƒ} }{\partial l} =  \frac{B²}{2 µ_{0}} N/m²

U-shaped magnet plus mass to be lifted

A_{C} = 12 cm² , air-gap length = 0.1 mm

Minimum force needd for lifting

=160 × g N

\left(\frac{B²}{2 µ_{0} }\right)  A_{C} = 160 g

B² = \frac{320  ×  9.81  ×  2  ×  4π  ×  10}{24  ×  10^{-4} }        or        B = 1.81 T

For B = 1.81 T,                        H = 2800 AT/m

Mean length of magnetic path = 75 cm

F (lifting magnet) = 2800 × 0.75 = 2100 AT

Air-gap length = 0.1 mm

A_{C} = 24 cm²

R_{g} = \frac{l}{μ_{0}  A } = \frac{0.1 × 10}{ 4π  ×  10^{-7}  ×  24  ×  10^{-4} }

= 33.2 × 10³

Flux

\phi _{g} = B A_{C} = 1.81 × 24 × 10^{-4}  = 4.344 × 10^{-3} Wb

A T _{g} = \phi_{g} R_{g} = 33.2 × 4.344

= 144

Total mmf needed

F = 2100 + 144 = 2244 AT

Minimum exciting current

i (min) = \frac{2244}{ 1000 } = 2.244 A