Question

A short magnet is arranged with its axis along east-west. A compass box is placed on its axial line at a distance of \[20cm\] from centre. If the deflection is \[{45^o}\]. Find moment of magnet. \[\left( {{B_H} = 0.4 \times {{10}^{ - 4}}T} \right)\]
a. \[0.8A{M^2}\]
b. \[1.6A{M^2}\]
c. \[2.4A{M^2}\]
d. \[3.2A{M^2}\]

Answer 1

Hint: For solving this question we must be aware of the concepts involved in this question. We know that the magnetic moment(M) is calculated by using a formula using magnetic field and distance and deflection of the short magnet as mentioned in the question above.

Formula used:
\[{B_{magnet}} = \dfrac{{{\mu _0}}}{{4\pi }}\dfrac{{2M}}{{{d^3}}}\]
Where, \[M\]- magnetic moment and \[d\]- distance to the axial line from the centre

Complete answer:
Here, a short magnet is arranged such that the axis of that magnet is along east-west and a compass box is placed on its axial line also there comes a deflection of \[{45^o}\], now we have to find the magnetic moment.

So, the magnetic field of box and the horizontal magnet is equal.
\[\therefore {B_{magnet}} = {B_H}\]

Therefore, we can say that
\[ \Rightarrow {B_H} = 0.4 \times {10^{ - 4}}T = \dfrac{{{\mu _0}}}{{4\pi }}\dfrac{{2M}}{{{d^3}}}\]
\[ \Rightarrow 0.4 \times {10^{ - 4}}T = \dfrac{{4\pi \times {{10}^{ - 7}}}}{{4\pi }}\dfrac{{2M}}{{{{(20 \times {{10}^{ - 2}}m)}^3}}}\]
\[ \Rightarrow \dfrac{{0.4 \times {{10}^{ - 4}}}}{{{{10}^{ - 7}}}} \times \dfrac{{{{\left( {0.2} \right)}^3}}}{2} = M\]
\[ \Rightarrow M = 0.0016 \times {10^3}A{M^2}\]
\[ \Rightarrow M = 1.6A{M^2}\]

So the answer is \[1.6A{M^2}\]. Option b is the correct answer.




Note: For more information we have to know that the magnetic field is equal to the horizontal compass’s magnetic field though there is deflection. The magnetic moment is calculated using the formula for magnetic field of the magnet.