Answer

Verified

404.7k+ views

**Hint:**Here, the copper rod is sliding under gravity. Due to the applied magnetic field, a force will be acting on the rod, and this force is proportional to the current in through the rod, its length, the applied magnetic field and the angle between magnetic field and copper rod. And a motional emf is also induced in the rod, which is proportional to the velocity of the conductor.

**Formula used:**

\[{{F}_{B}}=ilB\sin \theta \]

\[I=\dfrac{V}{R}\]

\[e=Bl{{v}_{T}}\]

**Complete answer:**

We have,

\[{{F}_{B}}=ilB\sin \theta \]

Where,

\[i\] is the current

\[l\] is the length of the conductor

\[B\] is the magnetic field

\[\theta \] is the angle between the rod and the magnetic field.

Here, magnetic field is acting perpendicular on the rod, \[\theta =90\]

Then,

Force,\[{{F}_{B}}=ilB\]

The force due to gravity is also acting downwards. Then, equating the horizontal component of forces at equilibrium,

\[mg\sin \theta -ilB=0\Rightarrow mg\sin \theta =ilB\] ---------- 1

We know that,

\[I=\dfrac{V}{R}\] -------------- 2

Where,

\[V\] is potential

\[R\] is resistance

Here, due the magnetic field, an emf is induced in the rod, then, potential will be equal to the induced emf.

\[V=e\]

Then, equation 2 becomes,

\[I=\dfrac{e}{R}\]

Substitute the above equation, in 1, we get,

\[mg\sin \theta =\dfrac{e}{R}lB\] ----------- 3

Since, the rod reached its terminal velocity, motional emf will be,

\[e=Bl{{v}_{T}}\]

Where,

\[{{v}_{T}}\]is the terminal velocity

Then,

Equation 3 becomes,

\[mg\sin \theta =\dfrac{Bl{{v}_{T}}}{R}\left( Bl \right)\Rightarrow {{v}_{T}}=\dfrac{\left( mg\sin \theta \right)R}{{{B}^{2}}{{l}^{2}}}\]

**Therefore, the answer is option B.**

**Note:**

Terminal velocity is a steady speed achieved by an object freely falling through a liquid or gas. An object released from rest will increase its speed until it attains the terminal velocity. When an object is forced to move faster than its terminal velocity, it will slow down to this constant velocity upon releasing.

Recently Updated Pages

How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE

Why Are Noble Gases NonReactive class 11 chemistry CBSE

Let X and Y be the sets of all positive divisors of class 11 maths CBSE

Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE

Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE

Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE

Trending doubts

Guru Purnima speech in English in 100 words class 7 english CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Which are the Top 10 Largest Countries of the World?

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

Difference Between Plant Cell and Animal Cell

Change the following sentences into negative and interrogative class 10 english CBSE

Select the word that is correctly spelled a Twelveth class 10 english CBSE

Give 10 examples for herbs , shrubs , climbers , creepers