Question

A uniform magnetic field of strength \( B = 2 \, \text{mT} \) exists vertically downwards. These magnetic field lines pass through a closed surface as shown in the figure. The closed surface consists of a hemisphere \( S_1 \), a right circular cone \( S_2 \), and a circular surface \( S_3 \). The magnetic flux through \( S_1 \) and \( S_2 \) are respectively

• A. \( \phi_1 = -20 \, \text{Wb}, \phi_2 = +20 \, \text{Wb} \)
• B. \( \phi_1 = +20 \, \text{Wb}, \phi_2 = -20 \, \text{Wb} \)
• C. \( \phi_1 = -40 \, \text{Wb}, \phi_2 = +40 \, \text{Wb} \)
• D. \( \phi_1 = +40 \, \text{Wb}, \phi_2 = -40 \, \text{Wb} \)

Answer 1

The Correct Option is A

Magnetic flux \( \phi \) is defined as the product of the magnetic field strength and the area through which the field lines pass, considering the direction of the field relative to the area:\( \phi = B \cdot A \cdot \cos(\theta) \)
For the magnetic flux through the hemisphere \( S_1 \) and the cone \( S_2 \), the direction of the magnetic field and the area of each surface are important. 
- The magnetic flux through the hemisphere is negative because the area vector is opposite to the direction of the magnetic field. 
- The flux through the cone is positive because the area vector and the magnetic field are aligned. 

Thus, the fluxes through the surfaces are:\( \phi_1 = -20 \, \text{Wb}, \quad \phi_2 = +20 \, \text{Wb} \)