Answer

Verified

96k+ views

**Hint**We will first establish a relationship between electric and magnetic fields at a point and then use it to calculate the magnitude of the magnetic field. The relation is given by:

\[\dfrac{{{B_0}}}{{{E_0}}} = C\]

Where,

\[{{\rm B}_0} = \] Magnetic Field at a point.

\[{{\rm E}_0} = \] Electric Field at a point.

c = speed of light \[\left( { = 3 \times {{10}^8}\dfrac{m}{s}} \right)\]

**Complete step by step solution:**

We will first determine the average energy at the point where we have to calculate the magnetic field value.

Average Energy in terms of magnetic field at a point is defined as:

\[{U_{av}} = \dfrac{1}{2}\dfrac{{{\rm B}_0^2}}{{{\mu _0}}}\]

Where,

\[{U_{av}} = \] Average Energy.

\[{\mu _0} = \] permeability of free space.

\[{B_0} = \] Magnetic Field at a point.

Average Energy in terms of magnetic field at a point is defined as:

\[{U_{av}} = \dfrac{1}{2}{\varepsilon _0}{\rm E}_0^2\]

Where,

\[{U_{av}} = \] Average Energy.

\[{\varepsilon _0} = \] permittivity of free space

\[{E_0} = \] Electric Field at a point.

Equating these two equations we get:

\[\dfrac{1}{2}\dfrac{{B_0^2}}{{{\mu _0}}} = \dfrac{1}{2}{\varepsilon _0}{\rm E}_0^2\]

\[\dfrac{{B_0^2}}{{{\mu _0}}} = {\varepsilon _0}E_0^2\]

\[\dfrac{{{\rm B}_0^2}}{{{\rm E}_0^2}} = {\mu _0}{\varepsilon _0}\]

Now using the relation, \[\mu {\varepsilon _0} = {c^2}\] where c = speed of light we can write:

\[\dfrac{{B_0^2}}{{E_0^2}} = {c^2}\]

\[ \Rightarrow \dfrac{{{B_0}}}{{{E_0}}} = c\]

In the question we have,

Electric Field \[({E_0}) = {10^{ - 3}}\dfrac{V}{m}\]

Speed of Light \[\left( c \right) = 3 \times {10^8}\dfrac{m}{s}\]

Magnetic Field \[({B_0}) = \] Required

Now, plugging in the values into the formula we get:

\[\dfrac{{{B_0}}}{{{{10}^{ - 3}}}} = 3 \times {10^8}\]

\[{B_0} = 3 \times {10^8} \times {10^{ - 3}}\]

\[ \Rightarrow {B_0} = 3 \times {10^5}\] Tesla

**Hence, option(d) is the correct solution.**

**Note:**In the question, the direction of the point which says ‘’At a point 10km due north of the transmitter’’ is extra information and has no use in the solution itself because the magnetic field is asked at the same point at which the value for peak electric field is given. Had these two points been different, then this information could have been of some use.

Recently Updated Pages

Write a composition in approximately 450 500 words class 10 english JEE_Main

Arrange the sentences P Q R between S1 and S5 such class 10 english JEE_Main

Write an article on the need and importance of sports class 10 english JEE_Main

Name the scale on which the destructive energy of an class 11 physics JEE_Main

Choose the exact meaning of the given idiomphrase The class 9 english JEE_Main

Choose the one which best expresses the meaning of class 9 english JEE_Main