Question

Consider a conical region of height h and base radius R with its vertex at the origin. Let the outward normal to its base be along the positive z-axis, as shown in the figure. A uniform magnetic field, \(\overrightarrow{B}=B_0\hat{z}\) exists everywhere. Then the magnetic flux through the base (\(\Phi_b\)) and that through the curved surface of the cone (\(\Phi_c\)) are

• A. \(\Phi_b=B_0\pi R^2;\Phi_c=0\)
• B. \(\Phi_b=-\frac{1}{2}B_0\pi R^2;\Phi_c=\frac{1}{2}B_0\pi R^2\)
• C. \(\Phi_b=0;\Phi_c=-B_0\pi R^2\)
• D. \(\Phi_b=B_0\pi R^2;\Phi_c=-B_0\pi R^2\)

Answer 1

The Correct Option is D

The problem involves calculating the magnetic flux through the base and the curved surface of a cone placed in a uniform magnetic field, \(\overrightarrow{B} = B_0 \hat{z}\).

1. Determine the magnetic flux through the base \(\Phi_b\):
   • The base of the cone is a circular plane with radius \(R\), and the area vector is along the positive z-axis. The magnetic field is uniform and also along the z-axis.
   • The area of the base is \(\pi R^2\).
   • The magnetic flux through the base is given by:
     \(\Phi_b = B_0 \times \text{Area of base} = B_0 \pi R^2\)
2. Determine the magnetic flux through the curved surface \(\Phi_c\):
   • The magnetic field is uniform and parallel to the z-axis.
   • The curved surface of the cone has no component of area perpendicular to the z-axis. This is because any elemental area on the curved surface has its normal perpendicular to the z-axis.
   • Therefore, the magnetic flux through the curved surface is:
     \(\Phi_c = 0\)

Using Gauss's Law for magnetism, we know that the net magnetic flux through a closed surface is zero. The flux through the base and the curved surface must sum to zero, verifying: \(\Phi_b + \Phi_c = 0 \Rightarrow B_0 \pi R^2 + 0 = 0\). However, the base flux is outward (positive), so the net result through the closed surface including open base and curved cone remains balanced.

Thus, the correct answer is:
\(\Phi_b = B_0 \pi R^2;\Phi_c = -B_0 \pi R^2\)