The magnetic induction at a distance of $20\,cm$ from centre of a short magnetic dipole having magnetic moment $0.5\,A{m^2}$ on its axis is:
A. $1.57 \times {10^{ - 5}}T$
B. $1.25 \times {10^{ - 5}}T$
C. $1.57 \times {10^{ - 4}}T$
D. $12.5 \times {10^{ - 5}}T$
Answer
527.1k+ views
Hint: In magnetism, as we know that magnetic moment is the strength of magnetic field produced by any material when placed in the presence of some magnetic field and magnetic induction is the magnitude of magnetic field produced by a body. We will use the basic relation between magnetic moment and magnetic induction to calculate the magnitude of magnetic induction.
Formula used:
If $B$ is the magnetic field at a distance of $d$ on axial line of a short magnetic dipole having magnetic moment of $m$ then
$B = \dfrac{{{\mu _0}2m}}{{4\pi {d^3}}}$
where, $\dfrac{{{\mu _0}}}{{4\pi }} = {10^{ - 7}}T$.
Complete step by step answer:
According to the question, we have given that the magnetic moment at a distance of $d = 20cm = 0.2m$ is $m = 0.5A{m^2}$ then, let B be the magnetic induction on axial line at given distance, then using the formula $B = \dfrac{{{\mu _0}2m}}{{4\pi {d^3}}}$ and putting the values of given parameters we get,
$B = \dfrac{{{\mu _0}2m}}{{4\pi {d^3}}}$
$\Rightarrow B = \dfrac{{{{10}^{ - 7}} \times 2 \times 0.5}}{{{{(0.2)}^3}}}$
On solving this calculation we get,
$B = \dfrac{{{{10}^{ - 7}}}}{{0.008}}$
$\therefore B = 1.25 \times {10^{ - 5}}\,T$
So, the magnetic induction at a distance of $20\,cm$ from centre of a short magnetic dipole having magnetic moment $0.5A{m^2}$ on its axis is $B = 1.25 \times {10^{ - 5}}\,T$.
Hence, the correct option is B.
Note:It should be remembered that, the basic unit of conversion is used in solving the question is $1\,cm = 0.001\,m$ and ${\mu _0}$ is known as the permeability of free space which has a fixed value of $4\pi \times {10^{ - 7}}T$ and here unit of magnetic induction is called Tesla denoted by $T$ and also magnetic moment is simply the current flowing in given surface area through a conductor.
Formula used:
If $B$ is the magnetic field at a distance of $d$ on axial line of a short magnetic dipole having magnetic moment of $m$ then
$B = \dfrac{{{\mu _0}2m}}{{4\pi {d^3}}}$
where, $\dfrac{{{\mu _0}}}{{4\pi }} = {10^{ - 7}}T$.
Complete step by step answer:
According to the question, we have given that the magnetic moment at a distance of $d = 20cm = 0.2m$ is $m = 0.5A{m^2}$ then, let B be the magnetic induction on axial line at given distance, then using the formula $B = \dfrac{{{\mu _0}2m}}{{4\pi {d^3}}}$ and putting the values of given parameters we get,
$B = \dfrac{{{\mu _0}2m}}{{4\pi {d^3}}}$
$\Rightarrow B = \dfrac{{{{10}^{ - 7}} \times 2 \times 0.5}}{{{{(0.2)}^3}}}$
On solving this calculation we get,
$B = \dfrac{{{{10}^{ - 7}}}}{{0.008}}$
$\therefore B = 1.25 \times {10^{ - 5}}\,T$
So, the magnetic induction at a distance of $20\,cm$ from centre of a short magnetic dipole having magnetic moment $0.5A{m^2}$ on its axis is $B = 1.25 \times {10^{ - 5}}\,T$.
Hence, the correct option is B.
Note:It should be remembered that, the basic unit of conversion is used in solving the question is $1\,cm = 0.001\,m$ and ${\mu _0}$ is known as the permeability of free space which has a fixed value of $4\pi \times {10^{ - 7}}T$ and here unit of magnetic induction is called Tesla denoted by $T$ and also magnetic moment is simply the current flowing in given surface area through a conductor.
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