JEE Main2013PhysicsElectromagnetic InductionActual
A circular loop of radius 0 . 3 cm lies parallel to a much bigger circular loop of radius 20 cm . The centre of the small loop is on the axis of the bigger loop. The distance between their centres is 15 cm . If a current of 2 . 0 A flows through the smaller loop, then the flux linked with a bigger loop is:
Options
- A3 .3 × 1 0 - 1 1 weber
- B6 .6 × 10 - 9 weber
- C9 .1 × 10 - 11 weber
- D6 × 10 - 11 weber
Correct answer
C. 9 .1 × 10 - 11 weber
Step-by-step solution
ϕ 2 = B 1 × A 2 = μ 0 I R 1 2 2 ( R 1 2 + x 2 ) 3 2 × π R 2 2 = μ 0 ( 2 ) ( 20 × 10 − 2 ) 2 2 [ ( 0.2 ) 2 + ( 0.15 ) 2 ] 3 2 × π ( 0.3 × 10 − 2 ) On solving = 9.216 × 1 0 - 1 1 ≈ 9.2 × 1 0 - 1 1  weber