99 Percentile Qs Bank for JEE MainPhysicsElectromagnetic Induction
A rectangular loop of wire is placed in the X Y -plane with its side of length 3 ~cm parallel to the X -axis and the side of length 4 ~cm parallel to the Y -axis. It is moving in the positive X -direction with the speed 10 ~cm / s . A magnetic field exists in the space with its direction parallel to the Z -axis. The field decreases by 2 10⁻³ ~T / cm along the positive X -axis and increases in time by 2 10⁻² ~T / s .
Options
- A-4.8 10⁻⁵ ~V
- B4.8 10⁻⁵ ~V
- C0
- D3.6 10⁻⁵ ~V
Correct answer
C. 0
Step-by-step solution
Induced emf in wire = Rate of change of flux aligned & = d d t (B A)=A ( d d t B )+v d d x B & =12 10⁻⁴ (2 10⁻²+10 10⁻² 2 10⁻³ ) & =12 10⁻⁴ (2 10⁻²+2 10⁻⁴ ) & =24 10⁻⁴ (10⁻²+10⁻⁴ )=24 10⁻⁵ ~V & =0.000024 ~V 0 ~V aligned