A bar magnet of magnetic moment $M$ is placed at right angles to a magnetic induction $B$. If a force $F$ is experienced by each pole of the magnet, the length of the magnet will be
A. $\dfrac{{MB}}{F}$
B. $\dfrac{{BF}}{M}$
C. $\dfrac{{MF}}{B}$
D. $\dfrac{F}{{MB}}$
Answer
564.6k+ views
Hint: The question states that a bar magnet is placed at right angles to a magnetic induction. So, we shall use the formula of torque experienced by a bar magnet when it interacts with the magnetic induction which is given as $\vec \tau = \vec M \times \vec B$ where $\tau $ is the torque, $M$ is the magnetic moment and $B$ is magnetic induction.Also, we will use the relation that expresses torque in terms of the force experienced by a body which is given as $\vec \tau = \vec r \times \vec F$ where $F$ is the force experienced by each pole of the magnet and $r$ is the distance from each pole from the point of rotation.We will then use the fact that the magnet and magnetic induction are placed at right angles. So, $\theta = {90^0}$ .Then we shall equate both the relations to get the answer.
Complete step by step answer:
The torque on a bar magnet with a given magnetic moment when placed in a magnetic induction is given as $\vec \tau = \vec M \times \vec B$ where $\tau $ is the torque, $M$ is the magnetic moment and B is magnetic induction.This can be rewritten as,
$\tau = MB\sin \theta $
Now it is given that $\theta = {90^0}$
$ \Rightarrow \tau = MB\sin {90^0}$
$ \Rightarrow \tau = MB\,\,\,\,\,\,\,\,\,\,.......(1)$
Now, the basic definition of torque says that it is the cross product of the force experienced at the point in consideration and the distance from the axis of rotation.
$\vec \tau = \vec r \times \vec F$
where $F$ is the force experienced by each pole of the magnet and $r$ is the distance from each pole from the point of rotation.
Hence, $r = L$
where $L$ is the length of the bar magnet.
$\vec \tau = \vec L \times \vec F$
This can be rewritten as $\tau = LF\sin \theta $
Now it is given that $\theta = {90^0}$
$ \Rightarrow \tau = LF\sin {90^0}$
$ \Rightarrow \tau = LF\,\,\,\,\,\,\,\,\,\,.......(2)$
From (1) and (2) we get,
$LF = MB$
$ \therefore L = \dfrac{{MB}}{F}$
Hence, option A is the correct answer.
Note:While dealing with cross products, the direction of every physical quantity in a particular physical relation must be known or derived. In this question, the direction of magnetic moment and magnetic induction was perpendicular. And since, the cross product always gives a perpendicular vector, all the three quantities-torque, magnetic moment and magnetic were mutually perpendicular. This makes it a special case and can be used directly in many numericals based on the same concept.
Complete step by step answer:
The torque on a bar magnet with a given magnetic moment when placed in a magnetic induction is given as $\vec \tau = \vec M \times \vec B$ where $\tau $ is the torque, $M$ is the magnetic moment and B is magnetic induction.This can be rewritten as,
$\tau = MB\sin \theta $
Now it is given that $\theta = {90^0}$
$ \Rightarrow \tau = MB\sin {90^0}$
$ \Rightarrow \tau = MB\,\,\,\,\,\,\,\,\,\,.......(1)$
Now, the basic definition of torque says that it is the cross product of the force experienced at the point in consideration and the distance from the axis of rotation.
$\vec \tau = \vec r \times \vec F$
where $F$ is the force experienced by each pole of the magnet and $r$ is the distance from each pole from the point of rotation.
Hence, $r = L$
where $L$ is the length of the bar magnet.
$\vec \tau = \vec L \times \vec F$
This can be rewritten as $\tau = LF\sin \theta $
Now it is given that $\theta = {90^0}$
$ \Rightarrow \tau = LF\sin {90^0}$
$ \Rightarrow \tau = LF\,\,\,\,\,\,\,\,\,\,.......(2)$
From (1) and (2) we get,
$LF = MB$
$ \therefore L = \dfrac{{MB}}{F}$
Hence, option A is the correct answer.
Note:While dealing with cross products, the direction of every physical quantity in a particular physical relation must be known or derived. In this question, the direction of magnetic moment and magnetic induction was perpendicular. And since, the cross product always gives a perpendicular vector, all the three quantities-torque, magnetic moment and magnetic were mutually perpendicular. This makes it a special case and can be used directly in many numericals based on the same concept.
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