Question

TA) Q4: Time (30 minutes) A. The current flowing in a solenoid, of 400 turns, 20 cm length & 4 cm diameter, changes with time according to the graph show to right. Derive an expression for the strength of the induced electric field inside the solenoid 0 00 01 02 03 04 Sketch the corresponding graph showing how the induced electric field vary with time. B. The Figure to right shows an infinite straight wire carries a current I is placed close to a rectangular loop of wire with width w and length l. i. Find the magnetic flux through the rectangular loop due to the current I. ii. What is the induced emf in the loop and the direction of the induced current if the current in the wire is changing in time such that (t) = a + bt, where a and bare constants?

Answer 1

2x olm au (A) total no. of twins N s 400 lengths ls 2004s diameter as 4CM - ragius &s 2cm = 2x10²m Inside the solenoid , Magnetic field B 3 C) 5 ew (A) I so total fury & BA I T22 Now fe.de de s dt ( / de la » E(297) solo 2 Es Mo & N 2 e di dt induced Electric field - 2 400 0.2 dI dt 57 substituting values 40X10 x 2 x 10 2 ob di at s -85 10 fo 5 s 50 A/S When & s oto O.is I charges from o to SA. = 1 -&T x 106 x 50 -6 - 400 x 105 V/m - 45 x 104 m So o s s d I so. for & s o.1 to 0.25 Is const. de for & s0.25 to o.us I decreases from sa to OA. 0-5 0.4-0.2 -5 s-25 Als so d dt 0.2

Sots-str x 106 x (-25) s 21 x 100 V/m . I 21 x 104 ECV/m) 0.4 0.2 f(s) -40x104 let take a 1 (B) for section of rectangular area of width da at distance a freom infinite streaight wire. Magnetic field B at x due I do I 2 B 02 2012 Plux to the da B d A Mo I de eda) Sin 212 Mo Il da S = = total of lux as S » & Mo Il [lax] 25 StW s 27 Mo Il 2T en (sth)

Ilds at but Induced emf es -da Mo Il dt 2 d I de 2t s en (sthi hol en (sth) a - Mol en (stwo Here are sign is for direction for Is atbt, induced and will to be created in such a way the direction of induced current will be Anticlockwise. 20 that, [] 100 Finduced.