JEE Main 2014 · 9 Apr · Shift 1 · Physics Q20

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PhysicsSingle correct+4 / -1Question 20 of 90

The magnetic field of earth at the equator is approximately $4\times10^{-5}$ T. The radius of earth is $6.4\times10^6$ m. Then the dipole moment of the earth will be nearly of the order of :

10^{23}

A m

^2

10^{20}

A m

^2

10^{16}

A m

^2

10^{10}

A m

^2

Answer and worked solution

Official answer: A

At the equator (the dipole's broadside position), $B=\dfrac{\mu_0}{4\pi}\dfrac{M}{R^3}$ :
$M=\frac{BR^3}{10^{-7}}=\frac{4\times10^{-5}\times(6.4\times10^6)^3}{10^{-7}}\approx1.0\times10^{23}\text{ A m}^2.$

Topics:Earth's magnetism

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JEE Main 2014 · 9 Apr · Shift 1, Physics question 20