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Question: The earth’s magnetic field at some place on the magnetic equator of the earth is \[0.5 \times {10^{ ...

The earth’s magnetic field at some place on the magnetic equator of the earth is 0.5×104T0.5 \times {10^{ - 4}}T. Consider the radius of the earth at the place as 6400 km. Then, the magnetic dipole moment of the earth is _____ Am2A{m^2}.(μ0=4π×107TmA1)\left( {{\mu _0} = 4\pi \times {{10}^{ - 7}}Tm{A^{ - 1}}} \right)
A. 1.31×10231.31 \times {10^{23}}
B. 1.05×10231.05 \times {10^{23}}
C. 1.15×10231.15 \times {10^{23}}
D. 1.62×10231.62 \times {10^{23}}

Explanation

Solution

In this question remember to use the concept of magnetic field also remember to use given values and substitute in the relation of magnetic field and magnetic dipole moment i.e. B=μ0M4π(Re)3B = \dfrac{{{\mu _0}M}}{{4\pi {{\left( {{R_e}} \right)}^3}}}, using this information will help you to approach the solution of the question.

Complete answer:
According to the given information it is given that, magnetic field at some place on magnetic equator is given 0.5×104T0.5 \times {10^{ - 4}}Twhere radius of earth is given as 6400 km
So, the given values are; B=0.5×104TB = 0.5 \times {10^{ - 4}}T, Re{R_e}= 6400 km or 6.4×1066.4 \times {10^6}m and μ0=4π×107TmA1{\mu _0} = 4\pi \times {10^{ - 7}}Tm{A^{ - 1}}
We know that by the relation of magnetic field and the magnetic dipole moment given as B=μ0M4π(Re)3B = \dfrac{{{\mu _0}M}}{{4\pi {{\left( {{R_e}} \right)}^3}}}where M is the magnetic dipole moment of the body, Re{R_e} is the radius of the earth, μ0{\mu _0}represents absolute magnetic permeability of free space and B is the magnetic field by the body
Substituting the given values in the relation of magnetic field and magnetic dipole moment we get
0.5×104=4π×107M4π×(6.4×106)30.5 \times {10^{ - 4}} = \dfrac{{4\pi \times {{10}^{ - 7}}M}}{{4\pi \times {{\left( {6.4 \times {{10}^6}} \right)}^3}}}
\Rightarrow M=0.5×104×262.14×1018107M = \dfrac{{0.5 \times {{10}^{ - 4}} \times 262.14 \times {{10}^{18}}}}{{{{10}^{ - 7}}}}
\Rightarrow M=131.07×1014107M = \dfrac{{131.07 \times {{10}^{14}}}}{{{{10}^{ - 7}}}}
\Rightarrow M=131.07×1021Am2M = 131.07 \times {10^{21}}A{m^2}
Or M=1.31×1023Am2M = 1.31 \times {10^{23}}A{m^2}
Therefore, magnetic dipole moment at the equator of the earth will be equal to 1.31×1023Am21.31 \times {10^{23}}A{m^2}

So, the correct answer is “Option A”.

Note:
In the above solution we used the term “magnetic field” which can be explained as the field under which the electric charge, electric current and magnetic objects experience the effect of magnetic force which acts perpendicular to the direction of velocity of charge, current or magnetic material the example of magnetic field is the magnetic force experienced by the magnetic materials on the earth where earth act as the source of magnetic field.