Question

A small bar magnet has a magnetic moment $1.2A{m^2}$. The magnetic field at a distance $0.1m$ on its axis will be:
$\left( {{\mu _0} = 4\pi \times {{10}^{ - 7}}Tm/A} \right)$

(A) $1.2 \times {10^{ - 4}}T$
(B) $2.4 \times {10^{ - 4}}T$
(C) $2.4 \times {10^4}T$
(D) $1.2 \times {10^4}T$

Answer 1

Hint:
First start with finding the relation between the magnetic moment and the magnetic field then put all the values from the question in that relation in order to get the required answer. Use the value of distance d and the ${\mu _0}$ in the relation and finally you will get the required answer.


Formula used :
magnetic field be B, is given by;
\[B = \dfrac{{{\mu _0}2M}}{{4\pi {d^3}}}\]



Complete step by step solution:

First start with the given information from the question:
Magnetic moment, $M = 1.2A{m^2}$
Distance, $d = 0.1m$
${\mu _0} = 4\pi \times {10^{ - 7}}Tm/A$
Let the magnetic field be B, is given by;
\[B = \dfrac{{{\mu _0}2M}}{{4\pi {d^3}}}\]
Putting all the values from the above, we get;
\[B = \dfrac{{4\pi \times {{10}^{ - 7}} \times 2 \times 1.2}}{{4\pi {{\left( {0.1} \right)}^3}}}\]
\[B = 2.4 \times {10^{ - 4}}T\]

Hence the correct answer is Option(B). \[\]






Note:
Use the same formula of the magnetic field no other formula will not be applicable here. Value of the all quantities was in the same unit and hence no need for unit conversion here, simply put all the values from the question in the formula of B and get the required answer.