Q150 / 160 · TS EAMCET 2022 (18 Jul Shift 1) · Physics

A planet has magnetic dipole moment of $27 \times 10^{22} A m^{2}$ . If the radius of the planet is $300 km$ , what would be the magnetic field at its equator?
$(Take, \frac{μ}{4 π} = 10^{- 7})$

Correct answer
A
$1 T$
B
$27 T$
C
$11 T$
D
$30 T$

Solution

Considering the whole planet as magnetic dipole, we can write magnetic field at equatorial position as,
$B_{e q u a t o r} = \frac{μ_{0}}{4 π} \times \frac{M}{r^{3}} = 10^{- 7} \times \frac{27 \times 10^{22}}{{(300 \times 10^{3})}^{3}} = 1 T$

Hence, the correct answer is option: 'A'.

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A planet has magnetic dipole moment of 27×1022 A m2. If the radius of the planet is 300 km, what would be the magnetic field at its equator?Take, μ4π=10-7

Solution

Relevant Questions

A planet has magnetic dipole moment of $27 \times 10^{22} A m^{2}$ . If the radius of the planet is $300 km$ , what would be the magnetic field at its equator?
$(Take, \frac{μ}{4 π} = 10^{- 7})$