of radius 
encloses a charge 
If there is another concentric sphere 
of radius 
and there is no additional change between and 
, then find the ratio of electric flux through 
and . 
of radius encloses a charge If there is another concentric sphere of radius and there is no additional change between and , then find the ratio of electric flux through and . Ans: We may recall that the expression for electric flux through a surface enclosing charge q by Gauss’s law is given by,
![\[\phi=\frac{q}{E_{0}}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-b39dfe258769d93f0df8e93ebd4f790b_l3.png "Rendered by QuickLaTeX.com")
Where, 
is the permittivity of the medium.
Now the electric flux through sphere is given by,
![\[\phi_{3_{1}}=\frac{Q}{e_{0}} \ldots \ldots(1)\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-1e2612c64e01a8d4cf3708183a01ffad_l3.png "Rendered by QuickLaTeX.com")
Since there is no additional charge between the given two spheres, the flux through sphere 
is given by,
We could now get the ratio of flux through spheres 5, and 
,
![\[\begin{array}{l} \frac{\phi_{s_{1}}}{\phi_{s_{r_{2}}}}=\frac{\frac{\mathrm{Q}}{\varepsilon_{\mathrm{D}}}}{\mathrm{Q}} \\ \therefore \frac{\phi_{\mathrm{S}_{\mathrm{S}}}}{\phi_{\mathrm{s}_{2}}}=\frac{1}{1} \end{array}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-faca46baf71fac11bbf4a058485d2ec3_l3.png "Rendered by QuickLaTeX.com")
Therefore, we find the required ratio to be 
.
.
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west of the geographic north. The north tip of the magnetic needle of a dip circle placed in the plane of magnetic meridian points above the horizontal. The horizontal component of the earth’s field is measured to be 
Specify the direction and magnitude of the earth’s field at the location.At a certain location in Africa, a compass points west of the geographic north. The north tip of the magnetic needle of a dip circle placed in the plane of magnetic meridian points above the horizontal. The horizontal component of the earth’s field is measured to be Specify the direction and magnitude of the earth’s field at the location.
and 
carrying currents of 
and 
in the same direction are separated by a distance of 
. Estimate the force on a 
section of wire A.
wire carrying a current of 
is placed inside a solenoid perpendicular to its axis. The magnetic field inside the solenoid is given to be 
. What is the magnetic force on the wire?
and making an angle of 
with the direction of a uniform magnetic field of 
. What is the magnitude of the field 
at a point 
from the wire?
-turns carrying current 
.. An electron moving at 
is at distance 
from the wire. Find the force acting on the electron if its velocity is directed towards the wire.