Answer

Verified

362.1k+ 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

Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred

Basicity of sulphurous acid and sulphuric acid are

The branch of science which deals with nature and natural class 10 physics CBSE

What is the stopping potential when the metal with class 12 physics JEE_Main

The momentum of a photon is 2 times 10 16gm cmsec Its class 12 physics JEE_Main

How do you arrange NH4 + BF3 H2O C2H2 in increasing class 11 chemistry CBSE

Trending doubts

How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE

What is the color of ferrous sulphate crystals? How does this color change after heating? Name the products formed on strongly heating ferrous sulphate crystals. What type of chemical reaction occurs in this type of change.

Write the difference between soap and detergent class 10 chemistry CBSE

Give 10 examples of unisexual and bisexual flowers

Differentiate between calcination and roasting class 11 chemistry CBSE

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

What is the difference between anaerobic aerobic respiration class 10 biology CBSE

a Why did Mendel choose pea plants for his experiments class 10 biology CBSE