A 2100-turn solenoid has a cross-sectional area equal to 8.0 cm2 and a length eq

Question

A 2100-turn solenoid has a cross-sectional area equal to 8.0 cm2 and a length equal to 30 cm. The solenoid carries a current of 4.0 A.
(a) Calculate the magnetic energy stored in the solenoid using U = ½ LI2, where L = µ0n2A.
____ mJ

(b) Divide your answer in part (a) by the volume of the region inside the solenoid to find the magnetic energy per unit volume in the solenoid.
____ kJ/m3

(c) Check your Part (b) result by computing the magnetic energy density from um = B2/(2µ0) where B = µ0nI.
_____ kJ/m3

Explanation / Answer

number of turns N = 2100 cross sectional area A = 8 cm2 = 8*10-4 m2 length l = 30 cm = 0.3 m current I = 4 A a) the magnetic energy stored in the solenoid is             U = (1/2)LI2    .............. (1) here , L = 0n2Al so , eq (1) becomes       U = (1/2)(0n2Al)I2              = (1/2)(0)(N/l)2(A)(l)(I)2   ................. (2) substitute the given data in eq (2) , we get         U = (1/2)(4*10-7 N/A2)(2100/0.3 m)2(8*10-4 m2)(0.3 m)(4 A)2             = 118.16*10-3 J            = 118.16 mJ   b) the magnetic energy per unit volume in the solenoid is U/V = U/Al        = (118.16*10-3 J) / (8*10-4 m2)(0.3 m)        = 0.492*103 J/m3        = 0.492 kJ/m3 c) the magnetic energy density is              u = (0N2I2) / (2)(l)2                     = [(4*10-7 N/A2)(2100)2 (4 A)2)] / (2)(0.3 m)2               = 0.492*103 J/m3               = 0.492 kJ/m3 length l = 30 cm = 0.3 m current I = 4 A a) the magnetic energy stored in the solenoid is             U = (1/2)LI2    .............. (1) here , L = 0n2Al so , eq (1) becomes       U = (1/2)(0n2Al)I2              = (1/2)(0)(N/l)2(A)(l)(I)2   ................. (2) substitute the given data in eq (2) , we get         U = (1/2)(4*10-7 N/A2)(2100/0.3 m)2(8*10-4 m2)(0.3 m)(4 A)2             = 118.16*10-3 J            = 118.16 mJ   b) the magnetic energy per unit volume in the solenoid is U/V = U/Al        = (118.16*10-3 J) / (8*10-4 m2)(0.3 m)        = 0.492*103 J/m3        = 0.492 kJ/m3 c) the magnetic energy density is              u = (0N2I2) / (2)(l)2                     = [(4*10-7 N/A2)(2100)2 (4 A)2)] / (2)(0.3 m)2               = 0.492*103 J/m3               = 0.492 kJ/m3             = 118.16*10-3 J            = 118.16 mJ   b) the magnetic energy per unit volume in the solenoid is U/V = U/Al        = (118.16*10-3 J) / (8*10-4 m2)(0.3 m)        = 0.492*103 J/m3        = 0.492 kJ/m3 c) the magnetic energy density is              u = (0N2I2) / (2)(l)2                     = [(4*10-7 N/A2)(2100)2 (4 A)2)] / (2)(0.3 m)2               = 0.492*103 J/m3               = 0.492 kJ/m3

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