KCET2024PhysicsElectromagnetic Induction
A uniform magnetic field of strength B=2 mT exists vertically downwards. These magnetic field lines pass through a closed surface as shown in the figure. The closed surface consists of a hemisphere S₁ , a right circular cone S₂ and a circular surface S₃ . The magnetic flux through S₁ and S₂ are respectively.
Options
- A_ S₁ =-20 ~Wb , _ S₂ =+20 ~Wb
- B_ S₁ =+20 ~Wb , _ S₂ =-20 ~Wb
- C_ S₁ =-40 ~Wb , _ S₃ =+40 ~Wb
- D_ S₁ =+40 ~Wb , _ S₂ =-40 ~Wb
Correct answer
A. _ S₁ =-20 ~Wb , _ S₂ =+20 ~Wb
Step-by-step solution
Given, B=2 mT =2 10⁻³ ~T , R= 10 ~cm Magnetic flux passing through surface S₁ , _ S₁ =B S₁ 180^ =2 10⁻³ R^2(-1)=-2 10⁻³ 100 10⁻⁴=-20 10⁻⁶ ~Wb =-20 ~Wb Since, Total entering magnetic flux = Total leaving magnetic flux _ S₂ =- _ S₁ =20 ~Wb