Question

The pole strength of a bar magnet is $48$ ampere-metre and the distance between its poles is $25cm$.The moment of the couple by which it can be placed at an angle of ${30^o}$ with the uniform magnetic intensity of flux density \[0.15{\text{ }}N{A^{ - 1}}{m^{ - 1}}\] will be
(A) $12Nm$
(B) $18Nm$
(C) $0.9Nm$
(D) None of the above

Answer 1

Hint:
To solve this question, we will first find the magnetization M and then use the general torque formula we will solve for the moment of couple required for the bar magnet to place it in a given condition.

Formula Used:

The magnetization M, the strength of magnet m, and length between poles L is related as $M = m.L$
Torque or moment of the couple is related as $\tau = MB\sin \theta $ where B is magnetic flux density and $\theta $ is the angle at which the bar magnet is placed.


Complete step by step solution:
According to the question, we have given that $m = 48Am$, $L = 25cm = 25 \times {10^{ - 2}}m$ so, magnetization is given by $M = m.L$ on putting the values, we get
$M = 48 \times 25 \times {10^{ - 2}}A{m^2} \to (i)$
now, the torque acting on the bar magnetic can be calculated using the formula $\tau = MB\sin \theta $ and we have given that $B = 0.15N{A^{ - 1}}{m^{ - 1}}$ and $\sin \theta = \sin {30^o} = 0.5$ on putting these values we get,
$\tau = MB\sin \theta $ also using value from equation (i) we have,
$
  \tau = 48 \times 25 \times 0.15 \times 0.5 \times {10^{ - 2}} \\
  \tau = 0.9Nm \\
 $
So, the moment of couple required to place the bar magnet is $0.9Nm$
Hence, the correct answer is option (C) $0.9Nm$




Therefore, the correct option is C.




Note:
It should be noted that in the formula of the moment of a couple $\tau = MB\sin \theta $, the magnetization and magnetic field density have the vector cross product and in vector form, it’s written as $\vec \tau = \vec M \times \vec B$, torque and moment of couples are just two words representing same meaning, they are not different.