Question

11. A solenoid with 120 turns over a length 3.0 mm has a current of 3.0 A. The figure below shows a cross-section of this solenoid. A small loop of wire (r-0.20 mm) is placed in the center of this solenoid. The small loop has a resistance of 1.5 m. (9 points) A) Draw the magnetic field from the solenoid.2 B) The solenoid's current began to increase by 2.0 A/s. Draw the direction of the induced magnetic field from the small loop. Label this vector.2 C) What is the induced current in the wire? B 120 003 BiMo (3.DA) 1ST I-E

Answer 1

Sorh) (6) we know that magnetic field inside a solenoid is given by Balo NI where N ý no. of turns ber unit length on Ba llon I where he is no e of turns and of O is length (in mj. so tada the magnetic field starts to increase as BaI so with in creasing magnetic field, the direction of induced magnetic field will be obposite to direction of original magnetic field.

Emf induced in dool is given by dt en la den E A demon I) > Allon dit we are given dI a 2.0 Al sec - dt and Putting Area = 192 = 1 (0 22X10-32 values we get e - (0.2X10-3² (419x10?) (120) 120 3x103 E = 6.32 x 108 Volta by so Induced current i i given ohn's law as I = € = 6.32X108 v 1.58603 II = 0.042 m A

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