
What is the terminal speed of the bar?
$ \left( A \right)\dfrac{{mgR}}{{{B^2}{l^2}\sin \phi }} \\
\left( B \right)\dfrac{{mgR\cos \phi }}{{{B^2}{l^2}{{\sin }^2}\phi }} \\
\left( C \right)\dfrac{{mgR\sin \phi }}{{{B^2}{l^2}{{\cos }^2}\phi }} \\
\left( D \right)\dfrac{{mgR\sin \phi }}{{{B^2}{l^2}\cos \phi }} \\ $
Answer
541.5k+ views
Hint :In order to solve the question, we are going to equate the magnetic field with the gravitational force, as in this condition only the bar will attain a terminal velocity. The terminal velocity is then derived from the equation by simplifying it and further solving the equation.
The formula used here is
Gravitational force is given by $ mg\sin \phi $
And magnetic force is given by $ Ilb\cos \phi $
Here, $ mg\sin \phi = Ilb\cos \phi $
Here $ m $ is the mass of the bar, $ g $ is the acceleration of gravity, $ B $ is the magnetic force, $ l $ is the length of the rod, $ b $ is the breadth, $ \phi $ is the angle.
The current in the bar $ I $ is given as
$ I = \dfrac{{Blv\cos \phi }}{R} $
Complete Step By Step Answer:
The bar attains the terminal velocity when the magnetic field is balanced by the gravitational force, so if we equate the two forces then, the equation becomes
$ mg\sin \phi = Ilb\cos \phi $
Here $ m $ is the mass of the bar, $ g $ is the acceleration of gravity, $ B $ is the magnetic force, $ l $ is the length of the rod, $ b $ is the breadth, $ \phi $ is the angle.
Where on left hand side, we have the gravitational force and on the right hand side, we have the magnetic field
The current in the bar $ I $ is given as
$ I = \dfrac{{Blv\cos \phi }}{R} $
Using this relation in the above equation
$ mg\sin \phi = \dfrac{{{B^2}{l^2}v{{\cos }^2}\phi }}{R} $
Rearranging the terms we get, the relation for velocity
$ v = \dfrac{{mgR\sin \phi }}{{{B^2}{l^2}{{\cos }^2}\phi }} $
Hence, the terminal speed of the bar $ v = \dfrac{{mgR\sin \phi }}{{{B^2}{l^2}{{\cos }^2}\phi }} $
Hence, option $ \left( C \right)\dfrac{{mg\sin \phi R}}{{{B^2}{l^2}{{\cos }^2}\phi }} $ is correct.
Note :
Terminal velocity is the steady speed that is achieved by the freely falling objects under gravity in air or in some medium. In this question, it is a very important step to see what all forces are acting on the bar and then checking the condition of equilibrium where the terminal velocity can be attained.
The formula used here is
Gravitational force is given by $ mg\sin \phi $
And magnetic force is given by $ Ilb\cos \phi $
Here, $ mg\sin \phi = Ilb\cos \phi $
Here $ m $ is the mass of the bar, $ g $ is the acceleration of gravity, $ B $ is the magnetic force, $ l $ is the length of the rod, $ b $ is the breadth, $ \phi $ is the angle.
The current in the bar $ I $ is given as
$ I = \dfrac{{Blv\cos \phi }}{R} $
Complete Step By Step Answer:
The bar attains the terminal velocity when the magnetic field is balanced by the gravitational force, so if we equate the two forces then, the equation becomes
$ mg\sin \phi = Ilb\cos \phi $
Here $ m $ is the mass of the bar, $ g $ is the acceleration of gravity, $ B $ is the magnetic force, $ l $ is the length of the rod, $ b $ is the breadth, $ \phi $ is the angle.
Where on left hand side, we have the gravitational force and on the right hand side, we have the magnetic field
The current in the bar $ I $ is given as
$ I = \dfrac{{Blv\cos \phi }}{R} $
Using this relation in the above equation
$ mg\sin \phi = \dfrac{{{B^2}{l^2}v{{\cos }^2}\phi }}{R} $
Rearranging the terms we get, the relation for velocity
$ v = \dfrac{{mgR\sin \phi }}{{{B^2}{l^2}{{\cos }^2}\phi }} $
Hence, the terminal speed of the bar $ v = \dfrac{{mgR\sin \phi }}{{{B^2}{l^2}{{\cos }^2}\phi }} $
Hence, option $ \left( C \right)\dfrac{{mg\sin \phi R}}{{{B^2}{l^2}{{\cos }^2}\phi }} $ is correct.
Note :
Terminal velocity is the steady speed that is achieved by the freely falling objects under gravity in air or in some medium. In this question, it is a very important step to see what all forces are acting on the bar and then checking the condition of equilibrium where the terminal velocity can be attained.
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