Electro Magnetic Induction And Alternating Currents Question 458
a conducting loop consisting of half circle of area $ A=0.06,m^{2} $ and three straight segments. The half circle lies in a uniform changing magnetic field $ B=4t^{2}+2t+5 $ (SI unit), where t is the time in second. An ideal battery E=2V is connected as shown and the total resistance of the wire is $ 2\Omega $ . The net current in the loop is at t=5 second is:
Options:
1A
B) 1.5A
C) 0.26A
D) 0.10A
Show Answer
Answer:
Correct Answer: C
Solution:
$ e_{ind}=\frac{d\phi }{dt}=\frac{d}{dt}(BA)=B\frac{dA}{dt} $
$ =A\frac{d}{dt}(4t^{2}+2t+5)=A(8t+2)=0.06(8t+2) $ =2.52 at t=5
$ \frac{i}{2} = 2 - e_{ind} = 0.26,A $
BETA
