Question

Earth's magnetic field at the equator 4 into 10 raise to minus 5 calculate a dipole moment​

Answer 1

Answer:

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Answer 2

ssuming that Earth's magnetic field is due to a bar magnet at the centre of the Earth, held along the polar axis of Earth, the equatorial magnetic field of Earth is given by

ssuming that Earth's magnetic field is due to a bar magnet at the centre of the Earth, held along the polar axis of Earth, the equatorial magnetic field of Earth is given by`B=(mu_0M)/(4pid^3)`

ssuming that Earth's magnetic field is due to a bar magnet at the centre of the Earth, held along the polar axis of Earth, the equatorial magnetic field of Earth is given by`B=(mu_0M)/(4pid^3)`where

ssuming that Earth's magnetic field is due to a bar magnet at the centre of the Earth, held along the polar axis of Earth, the equatorial magnetic field of Earth is given by`B=(mu_0M)/(4pid^3)`whered=R=Radius of the Earth

ssuming that Earth's magnetic field is due to a bar magnet at the centre of the Earth, held along the polar axis of Earth, the equatorial magnetic field of Earth is given by`B=(mu_0M)/(4pid^3)`whered=R=Radius of the Earthμ0=Absolute magnetic permeability of free space

ssuming that Earth's magnetic field is due to a bar magnet at the centre of the Earth, held along the polar axis of Earth, the equatorial magnetic field of Earth is given by`B=(mu_0M)/(4pid^3)`whered=R=Radius of the Earthμ0=Absolute magnetic permeability of free spaceM =Earth's dipole moment

ssuming that Earth's magnetic field is due to a bar magnet at the centre of the Earth, held along the polar axis of Earth, the equatorial magnetic field of Earth is given by`B=(mu_0M)/(4pid^3)`whered=R=Radius of the Earthμ0=Absolute magnetic permeability of free spaceM =Earth's dipole momentGiven:B=0.4 G=0.4×10−4 T

ssuming that Earth's magnetic field is due to a bar magnet at the centre of the Earth, held along the polar axis of Earth, the equatorial magnetic field of Earth is given by`B=(mu_0M)/(4pid^3)`whered=R=Radius of the Earthμ0=Absolute magnetic permeability of free spaceM =Earth's dipole momentGiven:B=0.4 G=0.4×10−4 T`mu_0/(4pi)=10^(-7) W_bA^(-1) m^(-1)`

ssuming that Earth's magnetic field is due to a bar magnet at the centre of the Earth, held along the polar axis of Earth, the equatorial magnetic field of Earth is given by`B=(mu_0M)/(4pid^3)`whered=R=Radius of the Earthμ0=Absolute magnetic permeability of free spaceM =Earth's dipole momentGiven:B=0.4 G=0.4×10−4 T`mu_0/(4pi)=10^(-7) W_bA^(-1) m^(-1)``d=6400 km=6400000 m=6.4×10^6 m`

ssuming that Earth's magnetic field is due to a bar magnet at the centre of the Earth, held along the polar axis of Earth, the equatorial magnetic field of Earth is given by`B=(mu_0M)/(4pid^3)`whered=R=Radius of the Earthμ0=Absolute magnetic permeability of free spaceM =Earth's dipole momentGiven:B=0.4 G=0.4×10−4 T`mu_0/(4pi)=10^(-7) W_bA^(-1) m^(-1)``d=6400 km=6400000 m=6.4×10^6 m``∴M=(Bd^3)/(10^(−7))`

ssuming that Earth's magnetic field is due to a bar magnet at the centre of the Earth, held along the polar axis of Earth, the equatorial magnetic field of Earth is given by`B=(mu_0M)/(4pid^3)`whered=R=Radius of the Earthμ0=Absolute magnetic permeability of free spaceM =Earth's dipole momentGiven:B=0.4 G=0.4×10−4 T`mu_0/(4pi)=10^(-7) W_bA^(-1) m^(-1)``d=6400 km=6400000 m=6.4×10^6 m``∴M=(Bd^3)/(10^(−7))``⇒M=((0.4xx10^(−4))(6.4xx10^6)^3)/10^(−7)`

ssuming that Earth's magnetic field is due to a bar magnet at the centre of the Earth, held along the polar axis of Earth, the equatorial magnetic field of Earth is given by`B=(mu_0M)/(4pid^3)`whered=R=Radius of the Earthμ0=Absolute magnetic permeability of free spaceM =Earth's dipole momentGiven:B=0.4 G=0.4×10−4 T`mu_0/(4pi)=10^(-7) W_bA^(-1) m^(-1)``d=6400 km=6400000 m=6.4×10^6 m``∴M=(Bd^3)/(10^(−7))``⇒M=((0.4xx10^(−4))(6.4xx10^6)^3)/10^(−7)`⇒M=1.05×1023 A m2