The net magnetic flux through any closed surface, kept in a magnetic field is

Question

The net magnetic flux through any closed surface, kept in a magnetic field is (A) zero (B) (μ0/4 π) (C) 4 π μ0 (D) (4 μ0/π)

Tardigrade Admin · Accepted Answer

According to Gauss's theorem in magnetism, surface integral of magnetic field intensity over a surface (closed or open) is always zero. <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/4775a05462495ee574d1904808da6446-.png" /> Number of field lines entering and going out from S are equal. So, net flux through S is zero. oint B ⋅ d A=0

Q. The net magnetic flux through any closed surface, kept in a magnetic field is

zero

$\frac{μ _{0}}{4 π}$

$4 π μ_{0}$

$\frac{4 μ _{0}}{π}$

According to Gauss's theorem in magnetism, surface integral of magnetic field intensity over a surface (closed or open) is always zero.

Number of field lines entering and going out from $S$ are equal. So, net flux through $S$ is zero.
$\oint B \cdot d A = 0$

Q. The net magnetic flux through any closed surface, kept in a magnetic field is

zero

4πμ0​​

4πμ0​

π4μ0​​

According to Gauss's theorem in magnetism, surface integral of magnetic field intensity over a surface (closed or open) is always zero. Number of field lines entering and going out from S are equal. So, net flux through S is zero. ∮B⋅dA=0

$\frac{μ _{0}}{4 π}$

$4 π μ_{0}$

$\frac{4 μ _{0}}{π}$

According to Gauss's theorem in magnetism, surface integral of magnetic field intensity over a surface (closed or open) is always zero.

Number of field lines entering and going out from $S$ are equal. So, net flux through $S$ is zero.
$\oint B \cdot d A = 0$