Solve any 10 problems:
Test 2 Fall, 2012
1.
In the above figure, the ideal batteries have emfs ε
1
=10 V and ε
2
=0.5 ε
1
, and the resistances are each 4 Ω. What is the current in (a) resistance 2 and (b) resistance 3 ?
Let current through R1 be I1 and through R2 be I2 as shown in figure. We will have
4I1+4(I1+I2)=10 or 8I1+4I2= 10 ----- (1)
Also 4I1+8I2=5 ---- (2)
Solving these equations, we get I1=5/4 and I2 = 0.
(a)Hence, current through R1 = 5/4=1.25.
(b) Current through R3 = 5/4=1.25A
2.
A 15 kΩ resistor and a capacitor are connected in series, and then a 12-V potential difference is suddenly applied across them. The potential difference across the capacitor rises to 5 V in 1.3 μs. (a) Calculate the time constant of the circuit. (b) Find the capacitance of the capacitor.
(a)We have
Vc

V


1
 e

 t


 

5 12 1
 e

1.3



 e

 t
1
5

7
12 12

1.3
7 ln
12

(b) RC=2.411*10^-6 giving C = 2.411*10^-6/15000
= 0.1607micro-farad
3.
A horizontal power line carries a current of 5000 A from south to north. Earth’s magnetic field (60 μT) is directed toward the north and inclined downward at 70 ͦ to the horizontal.
Find the (a) magnitude and (b) direction of the magnetic force on 100 m of the line due to
Earth’s field.

(a) Magnetic force = BILsinθ = 60*10^-6*5000*1*sin70 N/m = 0.2819N/m
(b) Direction is given by right hand rule. Thumb points in the direction of current and middle finger points in the direction of field. Thumb is pointing towards north and finger is at 70deg down to horizontal pointing towards north, then middle finger points towards west. Hence, force direction will be west.
4.
An electron in an old-fashioned TV camera tube is moving at 7.2x10
6 m/s in a magnetic field of strength 83 mT. What is the (a) maximum and (b) minimum magnitude of the force acting on the electron due to the field ? (c) At one point the electron has an acceleration of magnitude 4.9x10
14
m/s
2 . What is the angle between the electron’s velocity and the magnetic field ?
Force in magnetic field F= qBvsinθ
(a)
F will be maximum when θ=90 deg.
Hence, maximum force = 1.6*10
-19
*83*10
-6
*7.2*10^6 =9.5616*10
-14
N.
(b)
Minimum force will occur when θ=0 and; hence, F = 0.
(c) Acceleration f = F/m = qBvsinθ/m
Or,
 
19   
3   6
1.6 10 83 10 7.2 10 sin


31
  14 0.267
0
5.
In the above figure, two long straight wires are perpendicular to the page and separated by a distance d
1
=0.75 cm. Wire 1 carries 6.5 A into the page. What are the (a)magnitude and (b) direction (into or out of the page) of the current in wire 2 if the net magnetic field due to the two currents is zero at point P located at distance d
2
= 1.5 cm from wire 2 ?
Net field (H) at P =

I 1



I 2
2 ( 1 d 2) 2 d 2
0 I 2

6.5

1.5

4.33
A and current should be into the paper.

6.
The above figure shows a cross-section across a diameter of a long cylindrical conductor of radius a= 2 cm carrying a uniform current of 170 A. What is the magnitude of the current’s magnetic field at radial distance (a) 0, (b) 1 cm, (c) 2 cm (wire’s surface), and
(d) 4 cm ?
(a) B = 0 as current enclosed is zero.
(b) H

I
2
 r





1
2
170

2 2
2

0.01

678.408 /
  
0
H

4
 
10

7 
678.408

0.85
mT .
(c) H

I
2
 r

170
2
 
0.01

0
H

4
 
10

7 
170
2
 
0.01

3.4
mT .
(d) H

I
2
 r

170
2
 
0.04

0
H

4
 
10

7 
170
2
 
0.04

0.85
mT

7.
In the above figure, a wire forms a closed circular loop, with radius R=2cm and resistance 4 Ω. The circle is centered on a long straight wire ; at time t=0, the current in the long straight wire is 5 A rightward. Thereafter, the current changes according to i=
5A-(2 A/s
2
) t
2
. (The straight wire is insulated; so there is no electrical contact between it and the wire of the loop.) What is the magnitude of the current induced in the loop at time t > 0 ?
How current varies with time is to be correctly given?
8.
A long solenoid has a diameter of 12 cm. When a current I exists in the windings , a uniform magnetic field of magnitude B=30 mT is produced in its interior. By decreasing i, the field is caused to decrease at the rate of 6.5 mT/s . Calculate the magnitude of the induced electric field (a) 2.2 cm and (b) 8.2 cm from the axis of the solenoid.
(a) Flux linked
  r B EMF
1
  d
 dt
  r
1
2 dB dt
 
(2.2 / 100) 2   
3 
9.88

V
(b) Flux linked
 
2 r B

EMF
 d
 dt
  r
2 dB dt
 
(6 / 100)
2   
3 
73.51

V
9.
The current in an RL circuit drops from 1 A to 10 mA in the first second following removal of the battery from the circuit. If L is 10 H, find the resistance R in the circuit.
I

I e
0

R t
L 10 10

3
1 e
R
1
L
 e

R
L

0.01
R L ln 0.01
 
10 ln 0.01

4.605

10.
A coil is connected in series with a 10 kΩ resistor. An ideal 50 V battery is applied across the two devices, and the current reaches a value of 2 mA after 5 ms. (a) Find the inductance of the coil. (b) How much energy is stored in the coil at the same moment ?
I

I
0

1
 e

R t
L


2 10

3 
50
 3

1
 e

R t
L


 e

R t
L

0.6
L
(b)Energy stored = (1/2) LI
2
= 0.5*97.88*(2*10
-3
)^2 = 0.19576mJ
 ln 0.6

3

97.88
H

11.
Two coils are at fixed locations. When coil 1 has no current and the current in coil 2 increases at the rate 15 A/s, the emf in coil 1 is 25 mV. (a) What is their mutual inductance ? (b) When coil 2 has no current and coil 1 has a current of 3.6 A, what is the flux linkage in coil 2 ?
(a) Mutual inductance = e/(di/dt) = 25*10 -3 /15 = 1.67mH
(b) Flux linkage = mutual inductance*current = 1.67*10
-3
*3.6 = 0.006Wb-turns.
12.
A coil with 150 turns has a magnetic flux of 50 mT-m
2
through each turn when the current is 2 mA. (a) What is the inductance of the coil ? What are the (b) inductance and
(c) flux through each turn when the current is increased to 4 mA ? (d) What is the maximum emf ɛ across the coil when the current through it is given by i=(3 mA) cos(377t), with t in seconds ?
(a) Inductance = flux linkages/current = 150*50*10
-3
/2*10
-3
= 3750H
(b) Flux = inductance *current/turns = 3750*4*10
-3
/150=0.1T
(c)e =Ldi/dt = Ld(icoswt)/dt = Liw sinwt
Hence, maximum emf = Liw = 3750*3*10-3*377 = 4241.25volts.