A square coil of area 10^{-2} m^2 is placed | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The magnetic flux $\phi$ through a surface is given by the formula $\phi = B \cdot A \cdot \cos(	heta)$,where $B$ is the magnetic field intensity

Vedclass · Accepted Answer

$10$

Vedclass · Answer

(A) The magnetic flux $\phi$ through a surface is given by the formula $\phi = B \cdot A \cdot \cos(	heta)$,where $B$ is the magnetic field intensity,$A$ is the area of the coil,and $	heta$ is the angle between the magnetic field vector and the area vector.Since the coil is placed perpendicular to the magnetic field,the area vector (which is normal to the surface) is parallel to the magnetic field vector. Thus,$	heta = 0^\circ$ and $\cos(0^\circ) = 1$.Given: $B = 10^3 \ Wb/m^2$ and $A = 10^{-2} \ m^2$.Calculating the flux: $\phi = (10^3) 	imes (10^{-2}) 	imes \cos(0^\circ) = 10^1 = 10 \ Wb$.Therefore,the correct option is $A$.

$A$ square coil of area $10^{-2} \ m^2$ is placed perpendicular to a uniform magnetic field of intensity $10^3 \ Wb/m^2$. The magnetic flux through the coil is ........ $Wb$.

Similar Questions

$A$ loop,made of straight edges,has four corners at $A(L, L, 0)$,$B(-L, L, 0)$,$C(-L, -L, 0)$,and $D(L, -L, 0)$. $A$ magnetic field $\vec B = B_0(\hat i + \hat k) \text{ T}$ is present in the region. The magnetic flux passing through the loop $ABCD$ is:

The dimensional formula of magnetic flux is:

Verify Gauss's law for the magnetic field of a point dipole of dipole moment $\vec{m}$ at the origin for a surface which is a sphere of radius $R$.

Magnetic flux is in weber,the unit of magnetic field intensity is . . . . . . .

The unit of magnetic flux is

Vedclass Test Series

Exam Paper Generator

Online Exam Module