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Advanced Problems Of Motional Emf

Practice motional EMF equation with examples, calculate the force on a moving rod and velocity of rod as a function of time in a uniform magnetic field. And find velocity of rod as a function of time.

Force on a Moving Rod in a Uniform Magnetic Field

  • Consider a rod of mass m and length  \(\ell\) moving on two frictionless parallel rails in the presence of a uniform magnetic field directed into the page.
  • The rod is moving with velocity v as shown in figure.
  • As the bar slides, the current flows in counterclockwise direction into the circuit.
  • Due to this current, force \(F_B=I\ell B\)  is applied to the rod at the left side.
  • Force in horizontal direction

              \(F_x=I\ell B\) (Direction left side)

  • Force in vertical direction  

              \(F_y=\) Normal force (upward) + mg (downward)

  • The forces in y- direction cancel each other and only horizontal direction force remains.

            \(F_x=I\ell B\) (left side)

Note- The force is always opposite to \(v\) as it opposes the cause of change in flux according to Lenz's Law. 

Illustration Questions

Consider a rod of mass m and length \(\ell\), moving with velocity v inside the system of rail, as shown in figure. Find the force on the rod.

A \(I\ell B\,\hat j\)

B \(I\ell B\,\hat i\)

C \(I\ell B\,\hat k\)

D \(I\ell B\,(-\hat k)\)

×

As the bar slides, the current flows in clockwise direction. Due to this current, force \(F_B=I\ell B\) is applied on right side of rod.

\(F_B=I\ell B\,(\hat i)\)

image image

Consider a rod of mass m and length \(\ell\), moving with velocity v inside the system of rail, as shown in figure. Find the force on the rod.

image
A

\(I\ell B\,\hat j\)

.

B

\(I\ell B\,\hat i\)

C

\(I\ell B\,\hat k\)

D

\(I\ell B\,(-\hat k)\)

Option B is Correct

Terminal Velocity at which Rod Moves with Constant Velocity

  • Consider a system under gravity in which rod of mass m is moving with velocity v as shown in figure.

  • The magnetic force will be applied in upward direction which will balance gravitational force.

          \(F=mg\)

         \(I\ell B=mg\) ...(1)

  • Current  \(I\) is given as

           \(I=\dfrac{\mathcal{E}}{R}=\dfrac{B\ell v}{R}\)      \([\mathcal{E} =B\ell v]\)  ...(2)

  • From equation (1) and (2) 

        \(\dfrac{B\ell v(\ell B)}{R}=mg\)

         \(v=\dfrac{mgR}{B^2\ell^2}\)

  • This is the terminal velocity at which the net force will be zero and rod moves with constant velocity.

 

 

Illustration Questions

A conducting wire of length \(\ell=1\,m\), resistance \(R=2\,\Omega\) and mass \(m=1\,kg\), starts sliding at \(t=0\) down a smooth, vertical, thick pair of connected rails, as shown in figure. A uniform magnetic field \(B=1\,T\) exists in the space in a direction perpendicular to the plane of the rails. After sufficient time, the wire starts moving with constant velocity. Find this velocity.  \([g=10\,m/s^2]\)

A 20 m/sec

B 30 m/sec

C 40 m/sec

D 28 m/sec

×

The magnetic force will be applied in upward direction which will balance gravitational force.

          \(F=mg\)

          \(I\ell B=mg\) ...(1)

image

Current  \(I\) is given as

         \(I=\dfrac{\mathcal{E}}{R}=\dfrac{B\ell v}{R}\)      \([\mathcal{E} =B\ell v]\)  ...(2)

image

From equation (1) and (2) 

          \(\dfrac{B\ell v(\ell B)}{R}=mg\)

         \(v=\dfrac{mgR}{B^2\ell^2}\)

image

Given : \(m=1\,kg,\,g=10\,m/sec^2,\,R=2\,\Omega,\,B=1\,T,\,\ell=1\,m\)

\(v=\dfrac{1×10×2}{(1)^2×(1)^2}\)

\(v=20\,m/sec\)

image

A conducting wire of length \(\ell=1\,m\), resistance \(R=2\,\Omega\) and mass \(m=1\,kg\), starts sliding at \(t=0\) down a smooth, vertical, thick pair of connected rails, as shown in figure. A uniform magnetic field \(B=1\,T\) exists in the space in a direction perpendicular to the plane of the rails. After sufficient time, the wire starts moving with constant velocity. Find this velocity.  \([g=10\,m/s^2]\)

A

20 m/sec

.

B

30 m/sec

C

40 m/sec

D

28 m/sec

Option A is Correct

Velocity of Rod as a Function of Time in a Uniform Magnetic Field

Force on a moving rod in a uniform magnetic field

  • Consider a rod of mass m and length  \(\ell\) moving on two frictionless parallel rails in the presence of a uniform magnetic field directed into the page.
  • The rod is moving with velocity v as shown in figure.
  • As the bar slides, the current flows in counterclockwise direction into the circuit.
  • Due to this current, force \(F_B=I\ell B\) is applied to the rod at the left side.
  • Force in horizontal direction

              \(F_x=I\ell B\) (Direction left side)

  • Force in vertical direction 

             Fy = Normal force (upward) + mg (downward)

  • The forces in y- direction cancel each other and only horizontal direction force remains.

            \(F_x=I\ell B\) (left side)

Note- The force is always opposite to \(v\) as it opposes the cause of change in flux according to Lenz's Law. ?

\(F_x=-I\ell B\,(\hat i)\)

\(ma=\dfrac{-B^2\ell^2}{R}v(\hat i)\)   \(\left[I=\dfrac{\mathcal{E}}{R}=\dfrac{B\ell v}{R}\right]\)

\(m\dfrac{dv}{dt}=\dfrac{-B^2\ell^2}{R}v\)     \(\left[a=\dfrac{dv}{dt}\right]\)

\(m\dfrac{dv}{v}=\dfrac{-B^2\ell^2}{R}dt\)

\(\dfrac{dv}{v}=\dfrac{-B^2\ell^2}{mR}dt\)

Integrating both sides

\(\displaystyle\int\limits^v_{v_i}\dfrac{dv}{v}=\dfrac{-B^2\ell^2}{mR}\int\limits ^t_0dt\)    \(\left[At\;t=0,\,\text{initial velocity} =v_i,\\\ at \;t=t,\,\text{velocity}=v\right]\)

\(\ell n\left(\dfrac{v}{v_i}\right)=\dfrac{-B^2\ell^2}{mR}t\)

\(v=v_i\,e^{\dfrac{-B^2\ell^2}{mR}t}\)

\(v=v_i\,e^{-t/\tau}\,\hat i\)

where , \(\tau=\dfrac{mR}{B^2\ell^2}\)

 

 

Illustration Questions

Consider a rod that is given an initial velocity v in a system of rails having resistance R, in a region of magnetic field. To determine the velocity as a function of time, which of the following step is incorrect?

A As velocity is constant \(\therefore\,v(t)=v\)

B Find force on the rod to calculate acceleration \(F_x=I\ell B(\hat i)\)

C Put acceleration = \(\dfrac{dv}{dt}\)  \(m\dfrac{dv}{dt}=\dfrac{-B^2l^2v}{R}\)

D Integrate putting initial velocity and final velocity limits at time t=0 and t=t respectively \(\displaystyle\int\limits^v_{v_i}\dfrac{dv}{v}=\dfrac{-B^2\ell^2}{mR}\int\limits ^t_0dt\)  

×

Find the force on rod to calculate acceleration by right hand thumb rule.

Since, the velocity is a function of time, i.e., it varies with time, 

Then, acceleration  \(a=\dfrac{dv}{dt}\)

Integrate putting initial velocity and final velocity limits at time t = 0 and t = t respectively.

Hence, from above steps option (A) is incorrect.

Consider a rod that is given an initial velocity v in a system of rails having resistance R, in a region of magnetic field. To determine the velocity as a function of time, which of the following step is incorrect?

image
A

As velocity is constant

\(\therefore\,v(t)=v\)

.

B

Find force on the rod to calculate acceleration

\(F_x=I\ell B(\hat i)\)

C

Put acceleration = \(\dfrac{dv}{dt}\) 

\(m\dfrac{dv}{dt}=\dfrac{-B^2l^2v}{R}\)

D

Integrate putting initial velocity and final velocity limits at time t=0 and t=t respectively

\(\displaystyle\int\limits^v_{v_i}\dfrac{dv}{v}=\dfrac{-B^2\ell^2}{mR}\int\limits ^t_0dt\)

 

Option A is Correct

Energy Equation of a System

  • Consider the sliding rod as one system possessing kinetic energy which decreases because energy is transferring out of the rod by electrical transmission through the rails.
  • The resistor is considered as another system component possessing internal energy which rises as energy is transferred into it.
  • Since,

           Rate of energy transfer to resistor = Rate of energy transfer out of rod

         \(P_{resistor}=–P_{rod}\)

         \(I^2R=\dfrac{-d}{dt}\left(\dfrac{1}{2}mv^2\right)\)

        This is the energy equation of the system.

Illustration Questions

Consider a rod of mass m and length  \(\ell\) moving with velocity v on a system of rail as shown in figure. Choose the correct energy equation of a bar and a resistance.

A \(I^2R=\dfrac{1}{2}mv^2\)

B \(I^2R=\dfrac{d}{dt}\left(\dfrac{1}{2}mv^2\right)\)

C \(I^2R=\dfrac{-d}{dt}\left(\dfrac{1}{2}mv^2\right)\)

D \(I^2R+mv^2=0\)

×

Since,

          Rate of energy transfer to resistor = Rate of energy transfer out of rod

         \(P_{resistor}=–P_{rod}\)

         \(I^2R=\dfrac{-d}{dt}\left(\dfrac{1}{2}mv^2\right)\)

         This is the energy equation of the system.

Consider a rod of mass m and length  \(\ell\) moving with velocity v on a system of rail as shown in figure. Choose the correct energy equation of a bar and a resistance.

image
A

\(I^2R=\dfrac{1}{2}mv^2\)

.

B

\(I^2R=\dfrac{d}{dt}\left(\dfrac{1}{2}mv^2\right)\)

C

\(I^2R=\dfrac{-d}{dt}\left(\dfrac{1}{2}mv^2\right)\)

D

\(I^2R+mv^2=0\)

Option C is Correct

Terminal Velocity of a Rod on an Inclined Plane

  • A rod PQ of length \(\ell\), mass m and resistance R, slides on a smooth, thick pair of metallic rails joined at the bottom, as shown in figure.
  • The plane of rails is making an angle  \(\theta\) with the horizontal.
  • A vertical magnetic field B exists in the region.

  • Free body diagram of rod

  • Since \(v\,cos\theta\) is perpendicular to magnetic field B.

          \(I=\dfrac{B\ell \,v\,cos\theta}{R}\)

  • At terminal velocity net force is zero.

          \(F_{net}=0\)

  • Net force of y-direction

    The normal force in y-direction cancels \(mg\,cos\theta\) and \(F\,sin\theta\).

  • Net force in x-direction

    \(F\,cos\theta=mg\,sin\theta\)

    \(I\ell B\,\,cos\theta=mg\,sin\theta\)   \([F=I\ell B\ ]\)

    \(\dfrac{B\ell v\,cos\theta}{R}×\ell B \,cos\theta=mg\,sin\theta\)

    \(\Rightarrow\,\,\dfrac{B^2\ell^2\,cos^2\theta \,v}{1}=mgR\,sin\theta\)

    \(v_{terminal}=\dfrac{mgR\,sin\theta}{B^2\ell^2\,cos^2\theta}\)

    This is the terminal velocity of rod on an inclined plane.

Illustration Questions

A rod PQ of length \(\ell=1\,m\), mass  \(m=1\,kg\)  and resistance \(R=3\,\Omega\) on a smooth, thick pair of metallic rails, placed in a magnetic field  \(B=1\,T\). The plane of rails are making an angle \(\theta=45^\circ\) with the horizontal. Find the terminal velocity of rod PQ. \([g=10\,m/s^2]\) 

A \(5\,m/sec\)

B \(30\sqrt2\,m/sec\)

C \(7\,m/sec\)

D \(45\ m/sec\)

×

Free body diagram of rod

image

Since \(v\,cos\theta\) is perpendicular to magnetic field B.

          \(I=\dfrac{B\ell \,v\,cos\theta}{R}\)

At terminal velocity net force is zero.

           \(F_{net}=0\)

Net force of y- direction

The normal force in y-direction cancels \(mg\,cos\theta\) and \(F\,sin\theta\).

Net force in x-direction

\(F\,cos\theta=mg\,sin\theta\)

\(I\ell B\,\,cos\theta=mg\,sin\theta\)   \([F=I\ell B]\)

\(\dfrac{B\ell v\,cos\theta}{R}×\ell B \,cos\theta=mg\,sin\theta\)

\(\Rightarrow\,\,\dfrac{B^2\ell^2\,cos^2\theta \,v}{1}=mgR\,sin\theta\)

\(v_{terminal}=\dfrac{mgR\,sin\theta}{B^2\ell^2\,cos^2\theta}\)

This is the terminal velocity of rod on an inclined plane.

Given : \(\ell=1\ m,\,m=1\ kg,\,R=3\,\Omega,\,B=1\ T,\,\theta=45^\circ\)

\(v_{terminal} =\dfrac{1×10×3×sin45^\circ}{1×1×cos^245^\circ}\)

\(v_{terminal}=30\sqrt2\,m/sec\)

A rod PQ of length \(\ell=1\,m\), mass  \(m=1\,kg\)  and resistance \(R=3\,\Omega\) on a smooth, thick pair of metallic rails, placed in a magnetic field  \(B=1\,T\). The plane of rails are making an angle \(\theta=45^\circ\) with the horizontal. Find the terminal velocity of rod PQ. \([g=10\,m/s^2]\) 

image
A

\(5\,m/sec\)

.

B

\(30\sqrt2\,m/sec\)

C

\(7\,m/sec\)

D

\(45\ m/sec\)

Option B is Correct

Current in the Circuit placed in a Magnetic Field

  • Consider a rod of mass \(m\), and length \(\ell\) attached to a resistance \(R\) moving on two frictionless parallel wires in the presence of uniform electric field directed into the page.
  • The rod is moving with velocity \(v\) as shown in figure.
  • Due to motion of rod in magnetic field, force act such that positive charge accumulate a upper side of rod and negative charge at lower side of rod.
  • Thus it behave as battery as emf is induced due to motion of rod with positive terminal at upper end of rod and negative terminal at lower end.

         \(\mathcal{E} =B\ell v\)

 

  •  Equivalent circuit of the system is as shown in figure.
  • Current in the circuit is given as

       \(I=\dfrac{B\ell v}{R}\)

 

Illustration Questions

Consider a system in which two resistors are connected with a rod and moving with velocity v placed in a magnetic field \(\vec B\). Calculate the magnitude of current in the circuit. 

A \(\dfrac{\ell v}{B_1+B_2}\)

B \(\dfrac{B v}{R_1+R_2}\)

C \(\dfrac{v}{R_1+R_2}\)

D \(\dfrac{B\ell v}{R_1+R_2}\)

×

Equivalent circuit of the system

image image

Current in the circuit is given as

\(\mathcal{E}-IR_1-IR_2=0\)

\(\mathcal{E}=I(R_1+R_2)\)

\(I=\dfrac{\mathcal{E}}{R_1+R_2}\)

\(I=\dfrac{B\ell v}{R_1+R_2}\)   \([\mathcal{E}=B\ell v]\)

image

Consider a system in which two resistors are connected with a rod and moving with velocity v placed in a magnetic field \(\vec B\). Calculate the magnitude of current in the circuit. 

image
A

\(\dfrac{\ell v}{B_1+B_2}\)

.

B

\(\dfrac{B v}{R_1+R_2}\)

C

\(\dfrac{v}{R_1+R_2}\)

D

\(\dfrac{B\ell v}{R_1+R_2}\)

Option D is Correct

Current in a System Placed in a Magnetic Field

  • Consider two rods PQ and RS which are made to slide on the rails with the same speed v in a region of magnetic field, as shown in figure.
  • Both the rods are having same resistance R.
  • By using right hand rule, the direction of force can be determined which gives the direction of current.

  • Thus, the equivalent circuit is, as shown in figure.

  • Since both batteries are identical, so same current will flow from them. Let the current through them be \(I\).

         By applying loop,

       \(-IR+\mathcal{E}-2\,IR'=0\)

          \(I=\dfrac{\mathcal{E}}{R+2\,R'}\)

 

Illustration Questions

Consider two rods PQ and RS, each of length \(\ell=3\,m\)  and resistance \(R=2\,\Omega\) which are made to slide on the rails with the same speed \(v=2\ m/sec\) in a region of magnetic field \(B=1\,T\) with resistance \(R'=2\,\Omega\) connected across them. Find the current in R'.

A 1 A

B 4 A

C 2 A

D 3 A

×

By using right hand rule, the direction of force can be determined which gives the direction of current.

The equivalent circuit is, as shown in figure.

image

Since both batteries are identical, so same current will flow from them. Let the current through them be \(I\).

By applying loop,

\(-IR+\mathcal{E}-2\,IR'=0\)

\(I=\dfrac{\mathcal{E}}{R+2\,R'}\)

image

e.m.f of battery can be determined as 

\(\mathcal{E}=B\ell v\)

\(\mathcal{E}=1×3×2\)

\(\mathcal{E}=6\ V\)

Current across resistor R'

\(I=\dfrac{6}{2+2×2}\)

\(I=1\,A\)

 

 

Consider two rods PQ and RS, each of length \(\ell=3\,m\)  and resistance \(R=2\,\Omega\) which are made to slide on the rails with the same speed \(v=2\ m/sec\) in a region of magnetic field \(B=1\,T\) with resistance \(R'=2\,\Omega\) connected across them. Find the current in R'.

image
A

1 A

.

B

4 A

C

2 A

D

3 A

Option A is Correct

Velocity of Rod as a Function of Time in a Uniform Magnetic Field (using energy method)

  • Consider the sliding rod as one system possessing kinetic energy which decreases because energy is transferring out of the rod by electrical transmission through the rails.
  • The resistor is considered as another system component possessing internal energy which rises as energy is transferred into it.
  • Since,

          Rate of energy transfer to resistor = Rate of energy transfer out of rod

         \(P_{resistor}=–P_{rod}\)

          \(I^2R=\dfrac{-d}{dt}\left(\dfrac{1}{2}mv^2\right)\)

       This is the energy equation of the system.

  • We know \(I=\dfrac{B\ell v}{R}\)

          By energy equation 

           \(I^2R=\dfrac{-d}{dt}\,\left(\dfrac{1}{2}mv^2\right)\)

          \(\dfrac{B^2\ell^2v^2}{R}=\dfrac{-1}{2}m×2v\,\dfrac{dv}{dt}\)

         \(\dfrac{B^2\ell^2v^2}{R}=-mv\,\dfrac{dv}{dt}\)

  • Separating the variables

          \(\dfrac{dv}{v}=-\left(\dfrac{B^2\ell^2}{mR}\right)dt\)

  • At t=0, velocity is vi and at t=t, velocity is v.
  • Integrating both sides

          \(\displaystyle\int\limits^v_{v_i}\dfrac{dv}{v}=-\int\limits^t_{0}\dfrac{B^2\ell^2}{mR}\,dt\)

\(\Rightarrow\,\,\,\ell n\left(\dfrac{v}{v_i}\right)=-\left(\dfrac{B^2\ell^2}{mR}\right)t\)

\(\Rightarrow\,\,\,\dfrac{v}{v_i}=e^{-t/\tau}\)       where  \(\tau=\dfrac{mR}{B^2\ell^2}\)

\(\Rightarrow\,\,\,{v}={v_i}\,\,e^{-t/\tau}\)

 

Illustration Questions

Consider a rod that is given an initial velocity v in a system of rails, having resistance R, in a region of magnetic field. To determine the velocity as a function of time, which of the following step is incorrect?

A As velocity is constant, change in kinetic energy=0

B \(P_{resistor}=-P_{rod}\) \(I^2R=\dfrac{-d}{dt}\left(\dfrac{1mv^2}{2}\right)\)

C Differentiate kinetic energy of the rod \(\dfrac{B^2l^2v^2}{R}=-mv\dfrac{dv}{dt} \) where \(I=\dfrac{Blv}{R}\)

D Integrate by putting initial and final velocity of the rod at time t=0 and t=t respectively \(\displaystyle\int\limits^v_{vi}\dfrac{dv}{v}=-\int\limits^t_{0}\dfrac{B^2\ell^2}{mR}\,dt\)

×

\(P_{resistor}=-P_{rod}\)

\(I^2R=\dfrac{-d}{dt}\left(\dfrac{1mv^2}{2}\right)\)

Differentiate kinetic energy of the rod.

\(\dfrac{B^2l^2v^2}{R}=-mv\dfrac{dv}{dt} \)

where \(I=\dfrac{Blv}{R}\)

Integrate by putting initial and final velocity of the rod at time t=0 and t=t respectively.

\(\displaystyle\int\limits^v_{v_i}\dfrac{dv}{v}=-\int\limits^t_{0}\dfrac{B^2\ell^2}{mR}\,dt\)

Hence, from the following steps option (A) is incorrect.

Consider a rod that is given an initial velocity v in a system of rails, having resistance R, in a region of magnetic field. To determine the velocity as a function of time, which of the following step is incorrect?

image
A

As velocity is constant, change in kinetic energy=0

.

B

\(P_{resistor}=-P_{rod}\)

\(I^2R=\dfrac{-d}{dt}\left(\dfrac{1mv^2}{2}\right)\)

C

Differentiate kinetic energy of the rod

\(\dfrac{B^2l^2v^2}{R}=-mv\dfrac{dv}{dt} \)

where \(I=\dfrac{Blv}{R}\)

D

Integrate by putting initial and final velocity of the rod at time t=0 and t=t respectively

\(\displaystyle\int\limits^v_{vi}\dfrac{dv}{v}=-\int\limits^t_{0}\dfrac{B^2\ell^2}{mR}\,dt\)

Option A is Correct

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