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WBJEEA rigid wire loop of square shape having side of length L and...
A rigid wire loop of square shape having side of length L and resistance R is moving along the x-axis with a constant velocity v0 in the plane of the paper. At t = 0, the right edge of the loop enters a region of length 3L where there is a uniform magnetic field B0 into the plane of the paper, as shown in the figure. For sufficiently large v0, the loop eventually crosses the region. Let x be the location of the right edge of the loop. Let v(x), I(x) and F(x) represent the velocity of the loop, current in the loop, and force on the loop, respectively, as a function of x. Counter-clockwise current is taken as positive.
Which of the following schematic plot(s) is(are) correct ? (Ignore gravity)
Solution:
When loop was entering (x < L)
Ï•=BLx
e=dϕdt=-BLdxdt
|e|=BLV
i=eR=BLVR(ACW)
F=iB (Left direction) = B2L2VR (in left direction)
⇒ a=Fm=B2L2VmR a=VdVdx
VdVdx=B2L2VmR ⇒ V∫V0dV= -B2L2mR x∫0dx
⇒ V=V0-B2L2mRx (straight line of negative slope for x < L)
I=BLR V ⇒ (I vs x will also be straight line of negative slope for x < L)
L≤x≤3L
dϕdt=0
e=0 i=0
F = 0
x>4L
e=Blv
Force also will be in left direction.
i=BLWR (clockwise) a=B2L2VmR=VdVdx
F=B2L2VR x∫L-B2L2mRdx=Vf∫VidV
⇒ B2L2mR(x-L)=Vf-Vi
Vf=Vi-B2L2mR(x-L) (straight line of negative slope)
I=BLVR→ (Clockwise) (straight line of negative slope)
When loop was entering (x < L)
Ï•=BLx
e=dϕdt=-BLdxdt
|e|=BLV
i=eR=BLVR(ACW)
F=iB (Left direction) = B2L2VR (in left direction)
⇒ a=Fm=B2L2VmR a=VdVdx
VdVdx=B2L2VmR ⇒ V∫V0dV= -B2L2mR x∫0dx
⇒ V=V0-B2L2mRx (straight line of negative slope for x < L)
I=BLR V ⇒ (I vs x will also be straight line of negative slope for x < L)
L≤x≤3L
dϕdt=0
e=0 i=0
F = 0
x>4L
e=Blv
Force also will be in left direction.
i=BLWR (clockwise) a=B2L2VmR=VdVdx
F=B2L2VR x∫L-B2L2mRdx=Vf∫VidV
⇒ B2L2mR(x-L)=Vf-Vi
Vf=Vi-B2L2mR(x-L) (straight line of negative slope)
I=BLVR→ (Clockwise) (straight line of negative slope)
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