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.
\(F_x=I\ell B\) (Direction left side)
\(F_y=\) Normal force (upward) + mg (downward)
\(F_x=I\ell B\) (left side)
Note- The force is always opposite to \(v\) as it opposes the cause of change in flux according to Lenz's Law.
A \(I\ell B\,\hat j\)
B \(I\ell B\,\hat i\)
C \(I\ell B\,\hat k\)
D \(I\ell B\,(-\hat k)\)
\(F=mg\)
\(I\ell B=mg\) ...(1)
\(I=\dfrac{\mathcal{E}}{R}=\dfrac{B\ell v}{R}\) \([\mathcal{E} =B\ell v]\) ...(2)
\(\dfrac{B\ell v(\ell B)}{R}=mg\)
\(v=\dfrac{mgR}{B^2\ell^2}\)
\(F_x=I\ell B\) (Direction left side)
F_{y} = Normal force (upward) + mg (downward)
\(F_x=I\ell B\) (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 , \(\tau=\dfrac{mR}{B^2\ell^2}\)
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 \(\displaystyle\int\limits^v_{v_i}\dfrac{dv}{v}=\dfrac{-B^2\ell^2}{mR}\int\limits ^t_0dt\)
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.
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 \(I^2R+mv^2=0\)
\(I=\dfrac{B\ell \,v\,cos\theta}{R}\)
\(F_{net}=0\)
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.
A \(5\,m/sec\)
B \(30\sqrt2\,m/sec\)
C \(7\,m/sec\)
D \(45\ m/sec\)
\(\mathcal{E} =B\ell v\)
\(I=\dfrac{B\ell v}{R}\)
A \(\dfrac{\ell v}{B_1+B_2}\)
B \(\dfrac{B v}{R_1+R_2}\)
C \(\dfrac{v}{R_1+R_2}\)
D \(\dfrac{B\ell v}{R_1+R_2}\)
By applying loop,
\(-IR+\mathcal{E}-2\,IR'=0\)
\(I=\dfrac{\mathcal{E}}{R+2\,R'}\)
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.
By energy equation
\(I^2R=\dfrac{-d}{dt}\,\left(\dfrac{1}{2}mv^2\right)\)
\(\dfrac{B^2\ell^2v^2}{R}=\dfrac{-1}{2}m×2v\,\dfrac{dv}{dt}\)
\(\dfrac{B^2\ell^2v^2}{R}=-mv\,\dfrac{dv}{dt}\)
\(\dfrac{dv}{v}=-\left(\dfrac{B^2\ell^2}{mR}\right)dt\)
\(\displaystyle\int\limits^v_{v_i}\dfrac{dv}{v}=-\int\limits^t_{0}\dfrac{B^2\ell^2}{mR}\,dt\)
\(\Rightarrow\,\,\,\ell n\left(\dfrac{v}{v_i}\right)=-\left(\dfrac{B^2\ell^2}{mR}\right)t\)
\(\Rightarrow\,\,\,\dfrac{v}{v_i}=e^{-t/\tau}\) where \(\tau=\dfrac{mR}{B^2\ell^2}\)
\(\Rightarrow\,\,\,{v}={v_i}\,\,e^{-t/\tau}\)
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 \(\displaystyle\int\limits^v_{vi}\dfrac{dv}{v}=-\int\limits^t_{0}\dfrac{B^2\ell^2}{mR}\,dt\)