Question

8. A narrow metal bar with a resistance R and length l is constrained to slide along resistanceless metal rails, which are are connected on the right when the switch closes in either position A or position B, forming a complete loop of conducting material. A uniform magnetic field with magnitude |B| = Bo points into the page and occupies a region of length d. Ē pull R 0000 Le Ad ♡♡ B d a) For this part the switch is closed in position A (no current flows through the inductor L). Suppose the metal bar is pulled by a force Fpull to the left through the uniform magnetic field such that the speed vo of the metal bar is constant. Show that the work required to move the metal bar all the way through the magnetic field (so the bar travels the full distance d) at a constant speed equals the total energy dissipated in the metal bar (due to its resistance R). b) Now suppose we put the metal bar back where it started to the right of the magnetic field) and close the switch in position B so current can flow through the inductor. We again pull the metal bar through the uniform magnetic field at the same constant speed vo. Calculate the force required to maintain a constant speed as the bar moves through the magnetic field. Will this require more work, the same amount of work, or less work compared to part a)?

Answer 1

From Faraday's law of electromagnetic induction ,

we know that EMF (e ) induced in a moving conuctor in a magnetic field is given by , e = -d$\Phi$ / dt

$\Phi$ = magnetic flux which is equal to BA

where B = magnetic field , and A = area of the loop under magnetic field

The negative sign shows that the induced EMF (e) always opposes the net flux change.

( meaning that if the flux is increasing , then induced emf will ten to decrease the flux and vice versa)

Now,

8 Potential developed R will be Full IR R across e-Brl X X d

Taking the rod moved to a distance = x from initial point

Now Area closed by loop inide magnetic filed increase by , dA = l*x

[ where l = length of rod

x = length moved horizontally ]

now , d$\Phi$ /dt = d( BA)/dt

EMF (e) = B*[ dA/dt ] ( as B= B₀ = constant )

e = B*l [ dx/dt] ( as l = length is constant)

e = B₀lv₀ ( v₀ = velocity of rod )

Now as the conductor is moving with contant velocity which means that the net force acting on the rod should be zero. so definitely a force of same magnitude will act in the opposite direction due to induced emf

[ F_emf = force induced to the emf and acts in the right direction ]

as we know [ Force acting on a conductor rod moving in a magnetic field , F=I*L*B ]

  F_pull = F_emf = I*L*B ( I = current , L = length , B = magnetic field)

Now , In case a)

circuit is closed at A , containing only rod of resistance R.

As the force F_pull is pulling rod in the magnetic field therby increasing the net magnetic flux in the loop.

From ohm's law , I = V/R   

current flowing in the circuit is , I = e/R

I = B₀v₀l/R

now the energy dissipated in the resistor , is equal to the heat energy dissipated in the resistor .

W ( energy dissipated in resistor R) = I²Rt

where I = current

R = resistance of resistor

t = time

now putting the values of I = B₀v₀l/R ,

we get , W₁= (Bvl/R )²Rt

W₁ = [(B₀v₀l)²*t]/R

so W₁ = [(B_ol)²*v_o*(v_o*t) ]/R

W₁ = [(B_ol)_{​​​​​​}²*v_o*d]/R

( distance covered by rod moving with constant velocity is , d = v_o*t )

Now work done in moving rod will be , W₂ = F_pull * d [  work done = Force applied * distance moved]

= F_emf *d

= I*L*B*d => (Bvl/R)*l*B*d =>

W₂ = [(B₀l)² *v₀*d]/R --------(1)

so W1 = W2 , Hence proved.

b) in this case the loop is closed at B, so net resistance will be R_net = R + jXL   

X_L = reactive resistance of inductor

so , R_net = sqrt ( R² + (X_L)²) [ so R_net > R ]

now in this case current will be , i = e/R_net

[ e = same as previous one as it doesnt depend on the resistances in circuit]

Force required to move it with a constant velocty will be , F = i*l*B₀

  i = e/R_net

i = B₀v₀l/R_net

in this case the current induced will be less than the current induced in part a of the problem

Work done in moving the rod upto a distance of , d with constant velocity v₀ , W = F*d

W = F * d

W = (i*l*B)*d

W = [(B₀v₀l*l*B₀)*d] / R_net

so W = [(B₀l)²*v₀*d]/R_net   

comparing it to the work done in case a of the problem , by equation (1) we get that

only the denominator is different , so here R_net > R ( as calculated )

so Work done in case B will be less than the work done in case A