DEPARTMENT OF COMPUTER SCIENCE
MAT108B PHYSICS II
ASSIGNMENT 5
Question 1
The magnetic field is uniform and of magnitude B = 10 −4 T . The horizontal component of the field is
directly northward. The field also has a vertical component which is directed into the ground. The
angle the field lines dip below the horizontal is 45 o . A metal bar of length l = 1.5m carries a current of
I ' = 1 .2 A .
What is the magnitude of the magnetic force on the bar? Suppose
(a) The bar is held horizontally such that the current flows from East to West.
(b) The bar is held vertically such the current flows upward.
Answer:
( )
(a) F = I ' B sin 90 o = I ' B = (1.2 ) × 10 − 4 Nm −1.
The total force acting on the bar is
(
)
f = Fl = 1.2 × 10 − 4 (1.5) = 1.8 × 10 − 4 N
(b)
( )
F = I ' B sin 45 o = (1.2 ) 10 − 4 (0.707 ) = 8.48 × 10 − 5 Nm −1.
The total force acting on the bar is
f = Fl = 8.48 × 10 −5 (1.5) = 1.27 × 10 −4 N
(
)
Question 2
Suppose that an electron is accelerated from rest through a voltage difference of V = 102 volts and
then passes into a region containing a uniform magnetic field of magnitude B = 1.0T. The electron
subsequently executes a closed circular orbit in the plane perpendicular to the field. What is the radius
of this orbit? What is the angular frequency of gyration of the electron?
Answers:
If an electron of mass me = 9.11 × 10 −31 kg and charge e = 1.6 × 10 −19 C is accelerated from rest
1
through a potential difference V, then its final kinetic energy is me v 2 = eV .
2
The final velocity v of the electron is given by
v=
2eV
=
me
(
(2)(1.6 × 10 −19 )(10 2 ) = 5.93 × 10 6 ms −1
(9.11×10 − 31 )
)(
)
me v 9.11 × 10 −31 5.93 × 10 6
R=
=
= 3.38 × 10 −5 m
−19
eB
1.6 × 10 (1.0 )
ω=
(
(
)
)
eB 1.6 × 10 −19 (1.0 )
=
= 1.76 × 1011 rad .s −1
−
31
me
9.11 × 10
(
)
-15--
DEPARTMENT OF COMPUTER SCIENCE
MAT108B PHYSICS II
Question 3
A plane circular loop of conducting wire of radius r = 20cm which possesses N = 50 turns is placed in a
uniform magnetic field. The direction of the magnetic field makes an angle of 25o with respect to the
normal direction to the loop. The magnetic field-strength B is increased at a constant rate from B1 = 2 T
to B2 = 10 T in a time interval of ∆t = 15s . What is the emf generated around the loop? If the electrical
resistance of the loop is R = 20Ω , what current flows around the loop as the magnetic field is
increased?
Answer:
2
Area, A = πr 2 = π (0.2 ) = 0.1257 m 2
B⊥ = B cos θ = B cos 25 o = 0.906 B
The initial magnetic flux,
Φ B1 = NAB1 cos θ = (50)(0.1257 )(2)(0.906) = 11.39Wb
The final flux,
Φ B 2 = NAB 2 cos θ = (50)(0.1257 )(10 )(0.906) = 56.96Wb
dΦ B Φ B 2 − Φ B1 (56.96 − 11.39 )
=
= 3.038Wbs −1
dt
∆t
15
dΦ B
⇒ ε=
= 3.038V
dt
According to Ohm’s law, the current which flows around the loop in response to the emf is
ε (3.038)
I= =
= 0.152 A
R
(20)
Question 4
Two long, thin wires carry current in the positive y direction. Both wires are parallel to the y direction.
The 30A wire is in the z-y plane and is 3m from the y axis. The 10A wire is in the x-y plane and is 1 m
from the y axis. What is the magnitude of the magnetic field at the origin?
Answer:
30 A wire:
(µ I ) 4π × 10 − 7 Tm / A (30 A)
B1 = 0 =
= 2.00 × 10 − 6 T
(2π )(3m )
2πr
10 A wire:
(µ I ) 4π × 10 − 7 Tm / A (10 A)
B2 = 0 =
= 2.00 × 10 − 6 T
(2π )(1m )
2πr
The resultant field:
(
)
(
)
(
)
(
)
B1 cos 45 o + B2 cos 45 o = 2 2.00 × 10 − 6 T + 2 2.00 × 10 − 6 T = 2.83 × 10 − 6 T
Question 5
A − 8.0 µC charge is traveling at a speed of 8.0 × 10 6 m / s in a region of space where there is a
magnetic field. The angel between the velocity of the charge and the field is 45 o . A 5.0 × 10 −3 N force
r
acts on the charge. What is the magnitude of the B -field?
-16--
DEPARTMENT OF COMPUTER SCIENCE
MAT108B PHYSICS II
Answer:
F = q vB sin θ
B=
B=
F
q v sin θ
5.0 × 10 − 3 N
− 8.0 × 10
−6
(
6
)
C 8.0 × 10 m / s sin 45
o
= 1.1 × 10 − 4 T
Question 6
A long solenoid with an air core has n1 = 500 turns per meter and a cross-sectional area of A1 = 20cm2.
The current I1 flowing around the solenoid increases from 0 to 80 A in 4.0s. A plane loop of wire
consisting of n2 = 20 turns, which is of cross-sectional area A2 = 150cm2 and resistance R2 = 0.080Ω , is
placed around the solenoid close to its centre. The loop is orientated such that it lies in the plane
perpendicular to the axis of the solenoid. What is the magnitude ε 2 of the emf induced in the coil?
What current I2 does this emf drive around the coil?
Answer:
Φ B = N 2 A1 B
B = µ 0 n1 I1
⇒ Φ B = N 2 A1µ 0 n1 I1
dΦ B
dt
dI
= N 2 A1µ 0 n1 1
dt
(
)(
)
= (20 ) 20 × 10 − 4 4π × 10 − 7 (500 )
(80)
(4)
= 5.02 × 10 − 4 Wbs −1
By Faraday’s induction law,
dΦ B
ε2 = −
= −5.02 × 10 − 4 V
dt
Ohm’s law gives,
(
)
ε
− 5.02 × 10 − 4
I 2= 2 =
= −6.28mA
R2
(0.08)
ASSIGNMENT 6
Question 1
What is the quantity of charge on each capacitor if S2 is left open and S1 is closed as shown in Figure 1.
And how much charge will flow through the 12 Ω resistor? What effect would this have on the flow of
charge if the magnitude of resistor is double?
-17--
DEPARTMENT OF COMPUTER SCIENCE
MAT108B PHYSICS II
12V
7 µF
Figure 1
12 Ω
S1
3 µF
S2
1
1
1
=
+
C eq C1 C 2
Answer:
1
1
1
=
+
C eq 7 µF 3µF
C eq = 2.1µF
Q is the same for series.
Q = CV = (2.1µF )(12V ) = 25.2µC
The quantity of charge is 25.2 µC for each charge.
25.2 µC will flow through 12 Ω resistor.
The effect on the flow of charge is the same if the resistor is double.
Question 2
Examine the circuit in Figure 2 and calculate
(a) the potential difference across 3 µF capacitor,
(b) the charge on the plates of 2 µF capacitor,
(c) the energy associated with the charge stored in 4 µF capacitor.
6V
2 µF
4 µF
3 µF
Figure 2
Answer:
(a)
2 µF and 4 µF are connected in parallel
C eq1 = 2 µF + 4 µF = 6 µF
-18--
DEPARTMENT OF COMPUTER SCIENCE
MAT108B PHYSICS II
6 µF and 3 µF are connected in series
1
1
1
=
+
⇒ C eq 2 = 2 µF
C eq 2 6 µF 3µF
Q = CV = 2 µF (6V ) = 12 µC
Q is the same in series
So, 12 µC across 3 µF should be Veq1 =
Q
12µC
=
= 4V
C eq1
3µF
(b) V is the same in parallel
Therefore, 6V – 4V = 2 V across the 2 µF capacitor.
Q = CV = (2 µF) (2V) = 4 µC
(c) Q = CV = (4 µF) (2V) = 8 µC
V is the same in parallel
E=
1
1
QV = (8µC )(2V ) = 8µJ
2
2
Question 3
(a) What is the resistance of an 15 m aluminum wire with a radius 0.35 mm?
(b) If there is a potential difference of 0.5 V between the ends of the wire, what is the current flowing
through the wire? [Given that the resistivity of aluminum is 2.75 x 10-8 Ωm.]
Answer:
(a) By R = ρl/A, so
R = (2.75 x 10-8 Ωm) x (15 m) / π (0.00035 m)2
= 1.072 Ω
(b) From Ohm’s Law, V = IR
0.5 V = I (1.072Ω)
so the current through the wire is
I = 0.466 A
Question 4
Christmas tree lights are arranged in a circuit as shown in Figure 3. Each bulb dissipates 15W when
operated at 120 V.
(a) What is the resistance R of each bulb?
(b) What is the resistance of the entire array of bulbs?
(c) What is the total power consumption of the arrays?
(d) What are the currents at points a, b, c, and d?
-19--
DEPARTMENT OF COMPUTER SCIENCE
MAT108B PHYSICS II
R
R
c
d
R
R
R
R
R
R
b
a
120 V
Figure 3
Answer:
V2
V 2 (120V )2
⇒R=
=
= 960Ω
R
P
15W
1
1
1
1
1
1 1 1
1
1
1
1 1
(b)
= +
+ +
+ + = +
+ +
+ +
Rtotal R R + R R R + R R R R 2 R R 2 R R R
(a) P =
Rtotal =
R 960Ω
=
= 192Ω
5
5
V2
(120V ) = 75W
=
Rtotal
192Ω
2
(c) P =
(d) I a =
V
120V
=
= 0.625 A
Rtotal 192Ω
V 120V
=
= 0.125 A
R 960Ω
120V
I 2R =
= 0.0625 A
2 × 960Ω
I c = I a − 2 I b − 2 I 2 R = 0.625 − 2 × 0.125 − 2 × 0.0625 = 0.25 A
Ib =
I d = I b = 0.125 A
Question 5
Prove that when two resistors are connected in parallel, the equivalent resistance of the combination
is always smaller than that of either resistor.
Answer:
27-2:
 1
1 

Req =  +
 R1 R2 
−1
 R + R2 

=  1
 R1 R2 
−1
⇒ Req =
-20--
R1 R2
.
R1 + R21
DEPARTMENT OF COMPUTER SCIENCE
MAT108B PHYSICS II
⇒ Req = R1
R2
R1
< R1 and Req = R2
< R2 .
R1 + R2
R1 + R2
Question 6
(a) A battery with an emf of 12 V maintains a current of 0.6 A in an external circuit for 1 hr. What are
the power generated by the battery and the total energy expended by the battery?
(b) (i) Determine the currents I1, I2 and I3 in each branch of the circuit as shown in the Figure 4.
(ii) Calculate the power produced (or absorbed) by each source of emf.
(iii) Find the potential differences across the terminals of Battery 1, 2 and 3.
battery 1
I1
I3
1Ω
5Ω
12 V
A
BI3
battery 2
I2
I2
0.5 Ω
10 Ω
12 V
9.5Ω
battery 3
D
C
0.5Ω
15 V
Figure 4
Answer :
P = VI = (12V )(0.6 A) = 7.2W
60s
= 3600s
1 min
E = Pt = (7.2W )(3600s ) = 25.92kJ
t = 60 min
(b)(i)
For junction B,
I1
I 1 + I 3 = I 2 ……….(1)
For the upper loop of AB,
12 − (5Ω) I 1 − (10Ω) I 2 + 12 − (0.5Ω) I 2 − I 1 = 0
24 − 6 I 1 − 10.5I 2 = 0 ………...(2)
For the lower loop of ADCB,
15 − (10Ω) I 2 + 12 − (0.5Ω) I 2 − (9.5Ω) I 3 − (0.5Ω) I 3 = 0
27 − 10.5I 2 − 10I 3 = 0 ………...(3)
Sub (1) into (2),
-21--
DEPARTMENT OF COMPUTER SCIENCE
MAT108B PHYSICS II
24 − 6 I 1 − 10.5(I 1 + I 3 ) = 0
24 − 16.5 I 1 − 10.5 I 3 = 0 LL (4 )
Sub (1) into (3),
27 − 10.5(I 1 + I 3 ) − 10 I 3 = 0
27 − 10.5 I 1 − 20.5 I 3 = 0
27 − 10.5 I 1
LL (5)
20.5
Sub (5) into (4),
 27 − 10.5 I 1 
24 − 16.5 I 1 − 10.5
=0
20.5 

24 − 16.5 I 1 − 13.8 + 5.4 I 1 = 0
I3 =
I 1 = 0.92 A
27 − 10.5(0.92 )
= 0.85 A
20.5
I 2 = I 1 + I 3 = 0.92 + 0.85 = 1.77 A
I3 =
(ii)
For ε 1 ,
Power absorbed = ε 1 ⋅ I 1
= (12V )(0.92 A) = 11.04W
For ε 2 ,
Power absorbed = ε 2 ⋅ I 2
= (12V )(1.77 A) = 21.24W
For ε 3 ,
Power produced = ε 3 ⋅ I 3
= (15V )(0.85 A) = 12.75W
(iii)
For ε 1 , V1 = 12V − (1.0Ω)(0.92 A) = 11.08V
For ε 2 , V 2 = 12V − (0.5Ω)(1.77 A) = 11.12V
For ε 3 , V3 = 15.0V − (0.5Ω)(0.85 A) = 14.58V
ASSIGNMENT 7
Question 1
A capacitor is charged with 12.6 nC and has a 210 V potential difference between its terminals.
Compute its capacitance and the energy stored in it.
Answer:
Q 12.6 × 10 −9 C
C= =
V
210V
−11
C = 6 × 10 F = 60 pF
(
)
1
1
QV = 12.6 × 10 −9 C (210V )
2
2
E = 1.32 × 10 − 6 J = 1.32 µJ
E=
-22--
DEPARTMENT OF COMPUTER SCIENCE
MAT108B PHYSICS II
Question 2
Two capacitors in parallel, 3 µF and 5 µF, are connected, as a unit, in series with a 2 µF capacitor as
shown in Figure 1. The combination is connected across a 12 V battery. Find the equivalent
capacitance of the combination and the potential difference across the 3 µF capacitors.
3 µF
2 µF
5 µF
12 V
Figure 1
Answer:
3 µF and 5 µF are connected in parallel
C eq1 = 3µF + 5µF = 8µF
8 µF and 2 µF are connected in series
1
1
1
=
+
⇒ C eq 2 = 1.6 µF
C eq 2 8µF 2 µF
Q = CV = 1.6 µF (12V ) = 19.2 µC
Q is the same in series
So, 19.2 µC across Ceq1 should be Veq1 =
Q
19.2µC
=
= 2.4V
C eq1
8µF
V is the same in parallel
Therefore, 2.4 V across the 3 µF capacitor.
Question 3
(a) What is the resistance of an aluminum wire 12.0 m long with a diameter 0.8 mm?
(b) If there is a potential difference of 0.300 V between the ends of the wire, what is the current flowing
through the wire? [Given that the resistivity of aluminum is 2.65 x 10-8 Ωm.]
Answer:
(a) By R = ρl/A, so
R = (2.65 x 10-8 Ωm) x (12.0 m) / π (0.0004 m)2
= 0.633 Ω
(c) From Ohm’s Law, V = IR
0.300 V = I (0.633Ω)
so the current through the wire is
I = 0.474 A
-23--
DEPARTMENT OF COMPUTER SCIENCE
MAT108B PHYSICS II
Question 4
Six Christmas tree lights are arranged in a parallel circuit as shown in Figure 2. Each bulb dissipates
15W when operated at 120 V.
(e) What is the resistance R of each bulb?
(f) What is the resistance of the entire array of bulbs?
(g) What is the total power consumption of the arrays?
(h) What are the currents at points a, b, c, and d?
R
R
c
d
R
R
R
R
b
a
120 V
Figure 2
Answer:
V2
V 2 (120V )2
(a) P =
⇒R=
=
= 960Ω
R
P
15W
1
1 1 1 1 1 1 6
(b)
= + + + + + =
Rtotal R R R R R R R
Rtotal =
R 960Ω
=
= 160Ω
6
6
V2
(120V ) = 90W
(c) P =
=
Rtotal
160Ω
2
(d) I a =
V
120V
=
= 0.75 A
Rtotal 160Ω
V 120V
=
= 0.125 A
R 960Ω
I c = I a − 4 I b = 0.25 A
Ib =
I d = I b = 0.125 A
-24--
DEPARTMENT OF COMPUTER SCIENCE
MAT108B PHYSICS II
Question 5
Prove that when two resistors are connected in parallel, the equivalent resistance of the combination
is always smaller than that of either resistor.
Answer:
27-2:
 1
1 

Req =  +
 R1 R2 
⇒ Req = R1
−1
 R + R2 

=  1
 R1 R2 
−1
⇒ Req =
R1 R2
.
R1 + R21
R2
R1
< R1 and Req = R2
< R2 .
R1 + R2
R1 + R2
Question 6
(c) A battery with an emf of 9 V maintains a current of 0.5 A in an external circuit for 45 min. What are
the power generated by the battery and the total energy expended by the battery?
(d) (i) Determine the currents I1, I2 and I3 in each branch of the circuit as shown in the Figure 3.
(ii) Calculate the power produced (or absorbed) by each source of emf.
(iii) Find the potential differences across the terminals of Battery 1, 2 and 3.
battery 1
I1
I3
1Ω
5Ω
12 V
A
B
battery 2
I2
0.5 Ω
10 Ω
12 V
9.5Ω
battery 3
D
C
0.5Ω
15 V
Figure 3
Answer :
P = VI = (9V )(0.5 A) = 4.5W
60s
= 2700s
1 min
E = Pt = (4.5W )(2700s ) = 12.2kJ
t = 45 min
-25--
DEPARTMENT OF COMPUTER SCIENCE
MAT108B PHYSICS II
(b)(i)
For junction A,
I 1 + I 2 = I 3 ……….(1)
For the upper loop of ADCB,
(1.0Ω) I 1 + (5.0Ω) I 1 + (0.5Ω) I 3 + (9.5Ω) I 3 = 15V − 12V = 3V
6 I 1 + 10I 3 = 3 ………...(2)
For the lower loop of ADCB,
(0.5Ω) I 2 + (10.0Ω) I 2 + (0.5Ω) I 3 + (9.5Ω) I 3 = 15V − 12V = 3V
10.5I 2 + 10I 3 = 3 ………...(3)
Sub I 1 =
3 − 10 I 3
3 − 10 I 3
and I 2 =
into equation (1)
6
10.5
3 − 10 I 3 3 − 10 I 3
+
= I3
6
10.5
18 − 60 I 3 + 31.5 − 105 I 3 = 63I 3
228 I 3 = 49.5
I 3 = 0.217 A
3 − 10(0.217 )
= 0.138 A
6
3 − 10(0.217 )
I2 =
= 0.079 A
10.5
I1 =
(iv)
For ε 1 ,
Power absorbed = ε 1 ⋅ I 1
= (12V )(0.138 A) = 1.66W
For ε 2 ,
Power absorbed = ε 2 ⋅ I 2
= (12V )(0.079 A) = 0.95W
For ε 3 ,
Power produced = ε 3 ⋅ I 3
= (15V )(0.217 A) = 3.26W
(v)
For ε 1 , V1 = (1.0Ω)(0.138 A) + 12V = 12.14V
For ε 2 , V 2 = (0.5Ω)(0.079 A) + 12V = 12.04V
For ε 3 , V3 = 15.0V − (0.5Ω)(0.217 A) = 14.89V
ASSIGNMENT 8
Question 1
You want to produce a magnetic field with a magnitude of 6.7 × 10-4 T at a distance of 0.03 m from a
long, straight wire.
a)
What current is required to produce this field?
b)
With the current found in part (a), what is the magnitude of the field at a distance of 0.08 m from
the wire and at 0.16 m?
Answer:
-26--
DEPARTMENT OF COMPUTER SCIENCE
MAT108B PHYSICS II
29-12: a)
b)
B0 =
B=
2πrB0 2π (0.03 m)(6.7 × 10 −4 T )
µ0 I
⇒I=
=
= 100.5 A.
2πr
µ0
µ0
B × 0.03 (6.7 × 10 −4 T )(0.03)
µ0 I
, so B(r = 0.08 m) = 0
=
= 2.51 × 10 −4 T
2πr
0.08
0.08
B(r = 0.16 m) =
B0 × 0.03
= 1.26 × 10 − 4 T .
0.16
Question 2
Two long and fixed parallel wires, A and B, are 12 cm apart in air and carry currents of 50 A and 30 A,
respectively, in opposite directions. Determine the resultant flux density
(a) on a line midway between the wires and parallel to them, and
(b) on a line 8 cm from wire A and 20 cm from wire B.
IA = 50A
IB = 30A
8 cm
12 cm
Figure 1
Answer:
(a) At the midpoint between the wires, the fields both point into the page and hence reinforce:
µ (I + I ) 4π × 10−7 (50 + 30 )
B = BA + BB = 0 A B =
= 2.67 × 10− 4 T
2πr
2π (0.06)
(b) BA points out of the page and BB into the page.
I 
30 
µ I
 50
−5
B = BA − BB = 0  A − B  = 2 × 10− 7 
−
 = 9.5 × 10 T
2π  rA rB 
 0.08 0.2 
(
)
(
)
Question 3
r
A particle with charge 5.6 µC is moving with velocity v = −(3.5 × 10 3 m / s ) ˆj . The magnetic force on the
r
particle is measured to be F = + (6.8 × 10 −3 N )iˆ + (6.2 × 10 −3 N )kˆ .
(a) Calculate all the components of the magnetic field.
r r
r
r
(b) Calculate the scalar product B • F . What is the angle between B and F .
Answer:
27-9
(a)
-27--
DEPARTMENT OF COMPUTER SCIENCE
MAT108B PHYSICS II
r
r r
F = qv xB
(6.8 × 10
(6.8 × 10
−3
) (
) (
)[ (
) (
)
)]
) (
N iˆ + 6.2 × 10 −3 N kˆ = 5.6 × 10 −6 C − 3.5 × 10 3 m / s ˆj × B1iˆ + B2 ˆj + B3 kˆ
−3
N iˆ + 6.2 × 10 −3 N kˆ = (− 0.0196Cm / s )(B T ) − kˆ − (0.0196Cm / s )(B T ) iˆ
( )
1
3
()
6.8 × 10 −3 = −0.0196 B3 ⇒ B3 = −0.347
6.2 × 10 −3 = 0.0196 B1 ⇒ B1 = 0.316
r
B = 0.316iˆ − 0.347 kˆ
(b)
r r
r r
B • F = B F cos θ
(0r.316
iˆ − 0.347 kˆ )• (6.8 × 10
r
−3
)
(
)
(
)
iˆ + 6.2 × 10 −3 kˆ = (0.316 ) 6.8 × 10 −3 − (0.347 ) 6.2 × 10 −3 = 0
B F cos θ = 0
cos θ = 0
θ = 90°
Question 4
(a)
What is the force on a singly charged carbon ion (19.9x10-27 kg) moving with a speed of 3.2x105
m/s at right angles to a magnetic field of 0.84 T?
(b)
What is the centripetal acceleration of the ion?
(c)
What is the radius of the circle in which the ion moves?
Answer:
(a)
F = qBv
(
)
(
)
F = 1.6 × 10 −19 C (0.84T ) 3.2 × 10 5 m / s = 4.3 × 10 −14 N
(b)
F = ma ⇒ a =
(c)
r=
(
F
4.3 × 10 −14 N
=
= 2.16 × 1012 m / s 2
− 27
m 19.9 × 10 kg
)
2
v2
3.2 × 10 5 m / s
=
= 47.4mm
a 2.16 × 1012 m / s 2
Question 5
A coi1 3 cm in radius, containing 500 turns, is placed in a uniform magnetic field that varies with time
according to B = (0.015T / s )t + (3.2 × 10 −5 T / s 4 )t 4 . The coil is connected to a 450-Ω resistor, and its
plane is perpendicular to the magnetic field. The resistance of the coil can be neglected. a) Find the
magnitude of the induced emf in the coil as a function of time. b) What is the current in the resistor at
time t = 15 second?
Answer: [University Physics]
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DEPARTMENT OF COMPUTER SCIENCE
MAT108B PHYSICS II
30-6:
a)
NdΦ B
d
d
= NA ( B ) = (500 ) π 0.032
(0.015T / s )t + (3.2 x10− 5 T / s 3 )t 4
dt
dt
dt
(
ε=
) (
)
⇒ ε = 1.414 ((0.015 T/s) + (1.28 x 10-4 T/s3)t3)
= 0.021 V + (1.81 x 10-4 V/s3)t3.
b)
At t = 15 s ⇒ ε = 0.021 V + (1.81 x 10-4 V/s3(15 s)3 = +0.628V
⇒I=
ε
R
=
0.6280
= 1.4 × 10 −3 A.
450Ω
Question 6
A long, thin solenoid has 450 turns per meter an radius 1 cm. The current in the solenoid is increasing
di
. The induced electric field at a point near the center of the solenoid and 3.50 cm
dt
at a uniform rate
from its axis is 7.8 × 10-6 V/m. Calculate
di
.
dt
Answer: [University Physics]
30-24:
dΦ B
d
d
dI
dI E ⋅ 2πr
ε=
= ( BA) = ( µ0 .nIA) = µ0 nA ⇒
=
µ0 nA
dt
dt
dt
dt
dt
⇒
di (7.8 × 10 − 6V / m)(2π )(0.035m)
=
= 9.66 A / s.
dt
4π × 10 − 7 (450m −1 )π (0.01m) 2
(
)
or
dφ B
dt
turns 
dBA


−6 V 
 7.8 × 10
(0.035m ) = − 450
(0.035m )
m
m 
dt


ε = −N
7.8 × 10 −6
V
turns
= −450
(π )(0.01m)2 d  µ 0 i 
m
m
dt  2πr 
 4π × 10 −7 Wb

V
turns
2
A ⋅ m  di
(π )(0.01m) 
7.8 × 10
= −450
m
m
 2π × 0.035m  dt



−6 V 
 7.8 × 10
(2π )(0.035m )
di
m

=
2
dt
4π × 10 −7 Wb
− 450 turns π (0.01m )
A⋅m
m
= 9.66 A / s
−6
(
)(
)
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