- #1

lorenz0

- 148

- 28

- Homework Statement
- A conducting rod ##AC## of mass ##m=0.1\ kg##, length ##L=1\ m## and resistance ##R=5\ \Omega## can move without friction on two metallic rails in a region where there is a uniform magnetic field, perpendicular to the plane of the circuit and with magnitude changing with time.

##B(t_0=0)=2\ T## and the rod is in equilibrium at height ##h=3\ m##.

(a) Determine the magnetic force, the current (in direction and magnitude) and the electromotive force induced at time ##t_0=0##.

(b) Verify that if ##B(t)=B_0\sqrt{1-At}## the magnetic force is constant and find the value of ##A## such that the rod remains still.

(c) In these conditions, find the total charge that has gone around the circuit in the time interval from ##t_0=0\ s## to ##t_1=10\ s##.

(d) Find out for how much time the rod remains in equilibrium.

- Relevant Equations
- ##F_{m}=I\vec{L}\times\vec{B},\ \mathcal{E}=-\frac{d\phi(\vec{B})}{dt},\ V=IR##

I am having problems understanding point (b) so I would like to know if my reasoning in that part is correct and/or how to think about that part because I don't see how to justify the assumption ##v_y=0\ m/s##. Thanks.

I set up the ##xyz## coordinates system in the usual way with ##xy## in the plane and ##z## pointing upwards.

(a) The current at time ##0##, ##i(0)## should go around in a counterclockwise fashion otherwise it would help the weight pull the rod down and also ##i(0)LB(0)=mg\Rightarrow i(0)=\frac{mg}{LB(0)}## (1).

Also, by Ohm's Law we have that the electromotive force is ##\mathcal{E(0)}=i(0)R\overset{(1)}{=}\frac{mg}{LB(0)}##.

(b)

##i(t)=\frac{\mathcal{E(t)}}{R}=\frac{1}{R}(-\frac{d\phi(\vec{B})}{dt})=-\frac{1}{R}(\frac{d}{dt}\int_{S}\vec{B}\cdot d\vec{S})=-\frac{1}{R}(\frac{d}{dt}\int_{S}B_0\sqrt{1-At}dS)=-\frac{B_0}{R}(\frac{d}{dt}\sqrt{1-At}\int_S dS)=-\frac{B_0}{R}(\frac{d}{dt}\sqrt{1-At}\int_{0}^{y} L d\bar{y})=-\frac{LB_0}{R}(\frac{d}{dt}\sqrt{1-At} y)=-\frac{LB_0}{R}\left( -\frac{A}{2\sqrt{1-At}}\cdot y+\sqrt{1-At}\cdot v_y \right)##

so

##F_m(t)=i(t)LB(t)=-\frac{LB_0}{R}\left( -\frac{A}{2\sqrt{1-At}}\cdot y+\sqrt{1-At}\cdot v_y \right)\cdot L\cdot B_0\sqrt{1-At}=-\frac{L^2B_0^2}{R}\left( -\frac{A}{2}y+(1-At)v_y \right)##

so it seems to me that the only way that the magnetic force could be constant is if ##v_y=0## but can I assume this?

In that case ##F_m=\frac{L^2B_0^2 A}{2R}y## and to balance the weight it should be ##F_m-mg=0\Leftrightarrow \frac{L^2B_0^2 A}{2R}y=mg\Leftrightarrow A=\frac{2Rmg}{yL^2B_0^2}## but I am not sure which ##y## I should pick.

(c) ##q=\int_{t=0\ s}^{t=10\ s}i(t) dt##

I set up the ##xyz## coordinates system in the usual way with ##xy## in the plane and ##z## pointing upwards.

(a) The current at time ##0##, ##i(0)## should go around in a counterclockwise fashion otherwise it would help the weight pull the rod down and also ##i(0)LB(0)=mg\Rightarrow i(0)=\frac{mg}{LB(0)}## (1).

Also, by Ohm's Law we have that the electromotive force is ##\mathcal{E(0)}=i(0)R\overset{(1)}{=}\frac{mg}{LB(0)}##.

(b)

##i(t)=\frac{\mathcal{E(t)}}{R}=\frac{1}{R}(-\frac{d\phi(\vec{B})}{dt})=-\frac{1}{R}(\frac{d}{dt}\int_{S}\vec{B}\cdot d\vec{S})=-\frac{1}{R}(\frac{d}{dt}\int_{S}B_0\sqrt{1-At}dS)=-\frac{B_0}{R}(\frac{d}{dt}\sqrt{1-At}\int_S dS)=-\frac{B_0}{R}(\frac{d}{dt}\sqrt{1-At}\int_{0}^{y} L d\bar{y})=-\frac{LB_0}{R}(\frac{d}{dt}\sqrt{1-At} y)=-\frac{LB_0}{R}\left( -\frac{A}{2\sqrt{1-At}}\cdot y+\sqrt{1-At}\cdot v_y \right)##

so

##F_m(t)=i(t)LB(t)=-\frac{LB_0}{R}\left( -\frac{A}{2\sqrt{1-At}}\cdot y+\sqrt{1-At}\cdot v_y \right)\cdot L\cdot B_0\sqrt{1-At}=-\frac{L^2B_0^2}{R}\left( -\frac{A}{2}y+(1-At)v_y \right)##

so it seems to me that the only way that the magnetic force could be constant is if ##v_y=0## but can I assume this?

In that case ##F_m=\frac{L^2B_0^2 A}{2R}y## and to balance the weight it should be ##F_m-mg=0\Leftrightarrow \frac{L^2B_0^2 A}{2R}y=mg\Leftrightarrow A=\frac{2Rmg}{yL^2B_0^2}## but I am not sure which ##y## I should pick.

(c) ##q=\int_{t=0\ s}^{t=10\ s}i(t) dt##