• 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. 

• 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.

• 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.

• 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 v_i 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  \(\tau=\dfrac{mR}{B^2\ell^2}\)

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

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. ?

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

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

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

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

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

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 , \(\tau=\dfrac{mR}{B^2\ell^2}\)

• 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.

• 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

       \(I=\dfrac{B\ell v}{R}\)

• 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,

       \(-IR+\mathcal{E}-2\,IR'=0\)

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