Question

a selenoid is 20 cm long and has a cross sectional area of ​​5x10 ^ -4 m ^ 2. When the current through the selenoid decreases at a rate of 0.625 A / s, the induced emf is 2x10 ^ -4 V. Find the number of turns / unit length of the selenoid.

Timed Test This test has a time limit of 1 hour and 30 minutes. This test will save and submit automatically when the time expires Warnings appear when half the time 5 minutes, 1 minute, and 30 seconds remain. Multiple Attempts Not allowed. This test can only be taken once. Force Completion Once started, this test must be completed in one sitting Do not leave the test before clicking Save and Submit This test does not allow backtracking Changes to the answer after submission are prohibited Remaining Time: 30 minutes, 40 seconds. Less than half of the time remains Question Completion Status: » A Moving to the next question prevents changes to this answer Question 10 of 14 > Question 10 12 points Save Answer Un solenoide mide 20 cm de largo y tiene un área de sección transversal de 5x10-4 m2. Cuando la corriente a través del solenoide disminuye a razón de 0.625 A/s, la fem inducida es de 2x10-4 v. Hallar el número de vueltas/longitud unitaria del solenoide. O A 1825 OB 1325 OC 2050 OD 356 O E 725 OF 1596 A Moving to the next question prevents changes to this answer Question 10 of 14 > Type here to search CW O K! ENG 2:16 PM 7/31/2020

Answer 1

L = 200m =-0.204>Length of solenoid. A=5x104m2 cross sectional area of solenoid Let there are a number of tums per unit length. So, tatal number of turns will be (NxL). If current through the solenoid is a then, magnetic field inside it is B = Moni . Net magnetic flux through the solenoid is b= (NXL) BA > b=nx Lx Mon I A *16= n2xA MOLI magnitude of emf induced is do E= NYA MOL) d 1 sence, current decreases at the rate of 0.625A/S So, = 0.625 A/S = n dt dt dI at

2x104 n²x5x104 geven, E= 2x104N x40x1680.2X0:625 = 2.5x0X10-'xn »n²= 2x10-4 2 410-4 2.5X1 X10th = 2.54648X106 2.54648)x10 n=( sna na 1596 Correct option is