Question

A magnet has pole strength of $1000$ milli amp.m. The magnetic field intensity at a distance of $10cm$ from its North Pole is
(A) $\dfrac{{25}}{\pi }A/m$
(B) $\dfrac{\pi }{{25}}A/m$
(C) $\dfrac{{100}}{\pi }A/m$
(D) $\dfrac{\pi }{{100}}A/m$

Answer 1

Hint: To solve this question, we need to use the formula for the magnetic field due to a bar magnet. Then we have to use the relation between the magnetic field and the magnetic field intensity. Finally, putting the given values will give the final answer.

Formula used: The formulae used in solving this question are given by
$\Rightarrow B = \dfrac{{{\mu _0}}}{{4\pi }}\dfrac{m}{{{x^3}}}$, $B$ is the magnetic field produced due to a magnet of pole strength $m$ at a distance of $x$ from one of its poles.
$\Rightarrow H = \dfrac{B}{{{\mu _0}}}$, $B$ is the magnetic field, and $H$ is the magnetic field intensity.

Complete step by step answer
We know that the magnetic field due to a bar magnet at a distance of $x$ from one of its poles is given by
$\Rightarrow B = \dfrac{{{\mu _0}}}{{4\pi }}\dfrac{m}{{{x^3}}}$.......................(1)
Now, the magnetic field intensity is given by
$\Rightarrow H = \dfrac{B}{{{\mu _0}}}$
From (1)
$\Rightarrow H = \dfrac{1}{{4\pi }}\dfrac{m}{{{x^3}}}$.......................(2)
According to the question, we have $m = 1000mA - {\text{m}}$, and $x = 10cm$. Converting these into SI units, we get
$\Rightarrow m = 1000 \times {10^{ - 3}}A - {\text{m}}$
$ \Rightarrow m = 1A - {\text{m}}$.......................(3)
Also, $x = 10 \times {10^{ - 2}}m = 0.1m$.......................(4)
Substituting (3) and (4) in (2) we get
$\Rightarrow H = \dfrac{1}{{4\pi }}\dfrac{1}{{{{\left( {0.1} \right)}^2}}}$
$ \Rightarrow H = \dfrac{{100}}{{4\pi }}A/m$
On solving, we get
$\Rightarrow H = \dfrac{{25}}{\pi }A/m$
Thus, the magnetic field intensity is equal to $\dfrac{{25}}{\pi }A/m$.

Hence, the correct answer is option A.

Note
The formula for the magnetic field due to the bar magnet is similar to the expression for the equatorial electric field due to a dipole. We just need to replace the charge by pole strength, and the electrical permittivity in free space with the inverse of the magnetic permeability in the free space. Also, do not worry about the length of the magnet. The formula used in the above solution is derived after assuming the length to be negligible relative to the distance from the pole.