Gauss Law Of Magnetism MCQ - Practice Questions & Answers

Solve by difficulty

A square loop of side length 'a' is placed in a uniform magnetic field 'B' perpendicular to the plane of the loop. If the loop is rotated by the angle ' $\theta$ ' about an axis passing through its centre and parallel to one of its sides, what is the change in magnetic flux through the loop?

Zero

$a^2 B \cos \theta$

$-a^2 \cos \theta$

$a^2 B \cos \theta-a^2 B$

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Concepts Covered - 1

Magnetism and gauss's law

Magnetic flux (f)-

It is defined as the magnetic lines of force passing normally through a surface called magnetic flux.

As we learn in electrostatic, the Gauss law for a closed surface states that :

$\phi _{\text{closed}}=\frac{q_{net}}{\epsilon_{0} }$ ,

where $\phi =\int \bar{E} \dot d\bar{S}$

S is the area enclosed and E is the electric field intensity passing through it.

and $q_{net}$ is the total charge inside the closed surface.

But Gauss's Law of magnetism states that the flux of the magnetic field through any closed surface is zero (as shown in the below figure).

It is because inside the closed surface simplest magnetic element is a magnetic dipole with both the poles (since magnet with monopole does not exist). So a number of magnetic field lines entering the surface are equal to the number of magnetic field lines leaving the surface. So the net magnetic flux through any closed surface is zero.

Gauss's law in magnetism

I.e Gauss law for closed surface-

$\oint \underset{B}{\rightarrow}.\underset{ds}{\rightarrow} =0$

Gauss law if the surface is open

$\int \underset{B}{\rightarrow}.\underset{ds}{\rightarrow}=\phi _{B}$

i.e Consider an element of the area $dS$ on an arbitrarily shaped surface is shown in the figure. If the magnetic field at this element is $\vec B$ , the magnetic flux through the element is $d\phi_B=\vec B.d\vec S=BdS\cos\theta$