
The horizontal component of the flux density of earth’s magnetic field is $1.7 \times {10^{ - 5}}T$. The value of the horizontal component of the intensity of the earth’s magnetic field will be:
(A) $24.5A{m^{ - 1}}$
(B) $13.5A{m^{ - 1}}$
(C) $1.53A{m^{ - 1}}$
(D) $0.35A{m^{ - 1}}$
Answer
507k+ views
Hint
From the given value of the earth’s magnetic field, the field intensity can be found out by dividing the magnetic field with the permeability in free space. This can be given by the formula, $B = H{\mu _o}$.
Formula Used: In this solution, we will be using the following formula,
$B = H{\mu _o}$
where $B$ is the magnetic field,
$H$ is the magnetic field intensity and
${\mu _o}$ is the permeability in free space.
Complete step by step answer
Magnetic field is a vector field that is present in the vicinity of a magnet or a current-carrying conductor.
Our earth acts like a bar magnet having its north and south pole at the geographic north and south poles. It extends over a few thousand kilometers in space.
The magnetic field intensity is the part of the magnetic field in a material that arises from the external current and isn’t intrinsic to the material.
It is generally expressed by the vector $H$ and has a unit of $A{m^{ - 1}}$.
The magnetic field and the magnetic field intensity are related by the formula,
$B = H{\mu _o}$
From here we can write,
$ \Rightarrow H = \dfrac{B}{{{\mu _o}}}$
In the given question, the value of the magnetic field is given by, $B = 1.7 \times {10^{ - 5}}T$. And the value of the permeability in free space is given by, ${\mu _o} = 4\pi \times {10^{ - 7}}$. Therefore on substituting the values we get
$H = \dfrac{{1.7 \times {{10}^{ - 5}}}}{{4\pi \times {{10}^{ - 7}}}}$
On doing the calculation, we get the value,
$H = 13.5A{m^{ - 1}}$
So the correct answer is option B.
Note
The earth’s magnetic field is also called the geomagnetic field. It extends from the earth’s interior into space and there it interacts with the solar wind. The magnitude of the earth’s magnetic field on its surface ranges from 25 to 65 microteslas.
From the given value of the earth’s magnetic field, the field intensity can be found out by dividing the magnetic field with the permeability in free space. This can be given by the formula, $B = H{\mu _o}$.
Formula Used: In this solution, we will be using the following formula,
$B = H{\mu _o}$
where $B$ is the magnetic field,
$H$ is the magnetic field intensity and
${\mu _o}$ is the permeability in free space.
Complete step by step answer
Magnetic field is a vector field that is present in the vicinity of a magnet or a current-carrying conductor.
Our earth acts like a bar magnet having its north and south pole at the geographic north and south poles. It extends over a few thousand kilometers in space.
The magnetic field intensity is the part of the magnetic field in a material that arises from the external current and isn’t intrinsic to the material.
It is generally expressed by the vector $H$ and has a unit of $A{m^{ - 1}}$.
The magnetic field and the magnetic field intensity are related by the formula,
$B = H{\mu _o}$
From here we can write,
$ \Rightarrow H = \dfrac{B}{{{\mu _o}}}$
In the given question, the value of the magnetic field is given by, $B = 1.7 \times {10^{ - 5}}T$. And the value of the permeability in free space is given by, ${\mu _o} = 4\pi \times {10^{ - 7}}$. Therefore on substituting the values we get
$H = \dfrac{{1.7 \times {{10}^{ - 5}}}}{{4\pi \times {{10}^{ - 7}}}}$
On doing the calculation, we get the value,
$H = 13.5A{m^{ - 1}}$
So the correct answer is option B.
Note
The earth’s magnetic field is also called the geomagnetic field. It extends from the earth’s interior into space and there it interacts with the solar wind. The magnitude of the earth’s magnetic field on its surface ranges from 25 to 65 microteslas.
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