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# The transmitting antenna of a radio station is mounted vertically. At a point 10km due north of the transmitter the peak electric field is ${10^{ - 3}}\dfrac{V}{m}$ . The magnitude of the radiated magnetic field is:(a) $3.33 \times {10^{ - 10}}T$ (b) $3.33 \times {10^{ - 12}}T$ (c) ${10^{ - 3}}T$ (d) $3 \times {10^5}T$

Last updated date: 02nd Aug 2024
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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.