A conducting rod spans a gap of length L = 0.085 m and acts as the fourth side of a rectangular conducting loop, as shown in the figure. A constant magnetic field B = 0.55 T pointing into the paper is in the region. The rod is moving under an external force with an acceleration a = At, where A = 4,5 m/s*. The resistance in the wire is R = 145 N. L = 0.085 m B = 0.55 T A = 4.5 m/s R= 145 N X X х х X R V X X X Express the magnitude of the magnetic flux going through the loop, &, in terms of B, x and L. Express the speed of the rod, v, in terms of A and t. Assume v = 0 at t = 0. Express the position of the rod, x, in terms of A and t. Assume x = 0 at 1 = 0. %3= www
A conducting rod spans a gap of length L = 0.085 m and acts as the fourth side of a rectangular conducting loop, as shown in the figure. A constant magnetic field B = 0.55 T pointing into the paper is in the region. The rod is moving under an external force with an acceleration a = At, where A = 4,5 m/s*. The resistance in the wire is R = 145 N. L = 0.085 m B = 0.55 T A = 4.5 m/s R= 145 N X X х х X R V X X X Express the magnitude of the magnetic flux going through the loop, &, in terms of B, x and L. Express the speed of the rod, v, in terms of A and t. Assume v = 0 at t = 0. Express the position of the rod, x, in terms of A and t. Assume x = 0 at 1 = 0. %3= www
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![Ansuer-s
Given:
L= 0.085m
B = 0.55T
where. [A = 4.5mist]
R= 145
) Magastude ef flhux geing
KnaLo that, Magnctie flux is gien es
through the leep →
he
b= BA
6 = Magnetic feild
Area ef 'loop
where,
A
|0 = BaL
Here, [A =xL]
Step 3
b) Speed of Red
As Griven
accelevation l@)= A²
もaこ dv
.*, dv
dv= At*dt
both lides
Integ rating
Jdv =
V = A +C
At VE 0) & 0
: o +C
Thees, ue have
V= Af3
3
Step 4
c) Position rod >
we
Knoco theat
Vニ
ne have [v= Ad
.'. O becoes
dayat
こ
dr = At3 dt
Integrating bahsides –
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Transcribed Image Text:Ansuer-s
Given:
L= 0.085m
B = 0.55T
where. [A = 4.5mist]
R= 145
) Magastude ef flhux geing
KnaLo that, Magnctie flux is gien es
through the leep →
he
b= BA
6 = Magnetic feild
Area ef 'loop
where,
A
|0 = BaL
Here, [A =xL]
Step 3
b) Speed of Red
As Griven
accelevation l@)= A²
もaこ dv
.*, dv
dv= At*dt
both lides
Integ rating
Jdv =
V = A +C
At VE 0) & 0
: o +C
Thees, ue have
V= Af3
3
Step 4
c) Position rod >
we
Knoco theat
Vニ
ne have [v= Ad
.'. O becoes
dayat
こ
dr = At3 dt
Integrating bahsides –
=>
Transcribed Image Text:A conducting rod spans a gap of length L = 0.085 m and acts as the fourth side of
a rectangular conducting loop, as shown in the figure. A constant magnetic field B = 0.55 T pointing
into the paper is in the region. The rod is moving under an external force with an acceleration a =
At?, where A = 4,5 m/s*. The resistance in the wvire is R = 145 Q.
L = 0.085 m
B = 0.55 T
A = 4,5 m/s
R= 145 N
B
X
X
R
V
X
X
Express the magnitude of the magnetic flux going through the loop, , in terms of B, x and L.
Express the speed of the rod, v, in terms of A and t. Assume v = 0 at t = 0.
Express the position of the rod, x, in terms of 4 and t. Assume x = 0 at 1 = 0.
Express the derivative of the magnetic flux, do/dt, in terms of B, 4, L and t.
do/dt =
Express the magnitude of the emf induced in the loop, ɛ, in terms of B, L, A and t.
Express the current induced in the loop, I, in terms of ɛ and R.
Calculate the numerical value of I at t = 2s in A.
I(t=2s) =
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