1. A current of I is maintained in a single circular loop of circumference C. A magnetic
field of B is directed parallel to the plane of the loop.
(a) Calculate the magnetic moment of the loop.
(b) What is the magnitude of the torque exerted by the magnetic field on the loop?
Solution:
(a)
2 R  C  R 
C
2
2
1 C2
 C
A  R   
4 
 2 
1 IC2
  AI 
4 
2

(b)
B

 
    B     B sin 
    B sin 90o
  B

1 IBC2
4 
2. A magnetized sewing needle has a magnetic moment of . At its location, the Earth's
magnetic field is B northward at  below the horizontal. (a) Identify the orientation of
the needle that represent the minimum potential energy. (b) Identify the orientation of the
needle that represent the maximum potential energy of the needle–field system. (c) How
much work must be done on the system to move the needle from the minimum to the
maximum potential energy orientation?
Solution:

Horizon
B
(a) Potential energy is minimum when it is parallel to the magnetic field. i.e. the sewing
needle is pointing northward at  below the horizontal. U = - B cos 0o = -B in the case.
(b) Potential energy is maximum when it is anti-parallel to the magnetic field. i.e. the
sewing needle is pointing southward at  above the horizontal. U = B cos 0o = B in
the case.
(c) W = PE = Umax – Umin = B -(- B) = 2B
3. A rectangular coil consists of N closely wrapped turns and has dimensions a and b.
The coil is hinged along the y axis, and its plane makes an angle θ with the x axis (see
figure below).
(a) What is the magnitude of the torque exerted on the coil by a uniform magnetic field B
directed in the positive x direction when the current is I in the direction shown?
(b) If you are looking downward from the positive y direction, what is the expected
direction of rotation of the coil?
Solution:
(a)
Top view:
y


x

B
z

 
 
|  | |   B |  |  || B | sin 
 NIabB sin 
 NIabB sin (90o   )
 NIabB cos
(b)


y

z
x



  B

B
Using right hand rule, direction of
torque is given in the figure. If you
are looking downward from the
positive y direction, the torques is
in the –y direction (i.e. clockwise).
4. The rotor in a certain electric motor is a flat, rectangular coil with N turns of wire and
dimensions a by b. The rotor rotates in a uniform magnetic field of B. When the plane of the
rotor is perpendicular to the direction of the magnetic field, the rotor carries a current of I. In
this orientation, the magnetic moment of the rotor is directed opposite the magnetic field. The
rotor then turns through one-half revolution. This process is repeated to cause the rotor to turn
steadily at an angular speed of . (a) Find the maximum torque acting on the rotor. (b) Find
the peak power output of the motor. (c) Determine the amount of work performed by the
magnetic field on the rotor in every full revolution. (d) What is the average power of the
motor?
Solution:

B
(a )

 
 
|  ||   B |  |  || B | sin 

 is maximum when   90o and sin   1
 
  max |  || B |  NIabB
  
 

(b)     B  |  || B | sin  diection of  is perpendicular to the paper plane.

 is also perpendicular to the paper plane.
 
 P  |    | NIabB sin   NIabB sin t
P is maximum when   90o and sin   1
 Pmax  NIabB
T/2
(c) Amount of work done by the magnetic field in every reculution  2

NIabB sin t dt
0
 [-2NIabB cos t ]T/2
0
 (-2NIabB cos  ) - (-2NIabB cos 0)
 (2NIabB) - (-2NIabB)
 4NIabB
(d) Period T 
2

 Average power 
4NIabB 2NIabB

2


5. A Hall-effect probe operates with a current I. When the probe is placed in a uniform
magnetic field of magnitude B, it produces a Hall voltage of VH. (a) When it is used to
measure an unknown magnetic field, the Hall voltage is VH’. What is the magnitude of
the unknown field? (b) The thickness of the probe in the direction of B is t. Find the
density of the charge carriers, each of which has charge of magnitude e).
Solution:
(a)
Equation of motion :
V
 ev d B  0  V  d v d B
d
VH  d v d B
- -(1)
FE  FB  0  e

and VH '  d vd Bunknown
- -(2)
(2)/(1) 
VH '

VH
d v d Bunknown
d vd B

VH '

VH
Bunknown
B
 Bunknown 
VH '
B
VH
(b) VH  d v d B
But I  nev d A  v d 
I
neA
IB
IB
 I 
 n
 VH  d 
B 
VH et
net
 neA 
6. In an experiment designed to measure the Earth's magnetic field using the Hall effect, a
copper bar of thickness t is positioned along an east–west direction. Assume number of
electrons per cubic meter be n and the plane of the bar is rotated to be perpendicular to the
direction of B. If a current I in the conductor results in a Hall voltage of V, what is the
magnitude of the Earth's magnetic field at this location?
Earth’s magnetic
field
Equation of motion :
V
 ev d B  0  V  d v d B
d
I
But I  nev d A  v d 
neA
IB
Vnet
 I 
V  d 
 B
B 
net
I
 neA 
FE  FB  0  e
7. (a) In the figure below find the direction of the current in the wire that would produce a
magnetic field directed as shown.
Ans:
I
(b) In the figure below find the direction of the current in the wire that would produce a
magnetic field directed as shown.
Ans:
I
(c) In the figure below find the direction of the current in the wire that would produce a
magnetic field directed as shown.
Ans:
I
8. In Niels Bohr's 1913 model of the hydrogen atom, an electron circles the proton at a
distance of R with a speed of v. Compute the magnitude of the magnetic field this motion
produces at the location of the proton.
Solution:
v
eR
B=?
B
0I
2R
e
But I 
(T  period of the circular motion of the electron)
T
e
ev


2R 2R
v
 I 
 ev
ev
B  0  0 
 0 2
2R
2R 2R 4R
9. Consider the following figure.
(a) A conducting loop in the shape of a square of edge length  carries a current I as in the
figure above. Calculate the magnitude and direction of the magnetic field at the center of the
square. (b) If this conductor is reshaped to form a circular loop and carries the same current,
what is the value of the magnetic field at the center?
magnitude
Solution:
(a)
I A ds
r̂ 

 r
2
B
We just need to calculate the magnetic field due to segment AB and then multiply the
result with 8 for the squre.
Biot - Savart Law :


 0 I d s  r̂
 I ds sin 
 0
dB AB 
2
4 r
4
r2

d
But s  tan   ds 
2
2 cos 2


r 2 
cos  2cos 
  90 o    sin   sin (90 o   )  cos 
Substitute these resultes into (1) :
d
(
) cos 

 0 I 2 cos 2
dB AB 

4
(
)2
2cos 
- -(1)

 I 2d
 dB AB  0
cos 
4 

 I 2  /4
 B AB  0
cos d
4  0

 I
 B AB  0 [sin  ]0 / 4
2

 I 1
 B AB  0 (
 0)
2 2

0I
 B AB 
2 2

4 0 I
 Magnetic field at center due to the whole square  8B AB 
2

2 2 0 I

(b) If the conductor is reshaped to a circular loop, circumference = 4 
 2R  4  R 
B
0I
2R
2

0I
 B
2(
 B
2
)

 0 I
4
10. A current path shaped as shown in the figure produces a magnetic field at P, the
center of the arc. If the arc subtends an angle of θ and the radius of the arc is R, what are
the magnitude and direction of the field produced at P if the current is I?
Solution:
Magnetic field due to the two straight currents = 0 (why?)
Magnetic field due to the arc 
0I
2R
 I
 0
4R
 I
 Magnetic field at point P  0
4R
 I
 0
4R


2
00
11. Two long, parallel wires carry currents of I1 and I2 in the directions indicated in the figure
below, where d is the distance between the two wires. Take the positive x direction to be to
the right.
(a) Find the magnitude and direction of the magnetic field at a point midway between the
wires. (b) Find the magnitude and direction of the magnetic field at point P, located d
above the wire carrying the current I2.
Solution:
B1
I1
(a)
I2
d/2

 I
B1  0 1 ŷ 
2 r1
 0 I1
ŷ 
B2
d/2
 0 I1
ŷ
d
d
2 ( )
2

 I
 I
 I
B 2   0 2 ŷ  - 0 2 ŷ  - 0 2 ŷ
d
2 r1
d
2 ( )
2
 

 I
 I
 B  B1  B 2  0 1 ŷ - 0 2 ŷ
d
d

0
(I1 - I 2 )ŷ
d
y
x
B1
(b)
B2
45o
r1
I1
r2=d
I2
o
45
y
x
d

 I
| B1 |  0 1
2 r1
But r1  d 2  d 2  2 d

 0 I1
 | B1 | 
2 2 d



B1  - | B1 | cos 45 o x̂  | B1 | cos 45 o ŷ
 0 I1
 0 I1
1
1


ŷ
x̂ 
2 2 d 2
2 2 d 2
 I
 I
 - 0 1 x̂  0 1 ŷ
4 d
4 d

 I
 I
B 2   0 2 x̂  - 0 2 x̂
2 d
2 r1
 

 I
 I
 I
 B  B1  B 2  (- 0 1 x̂  0 1 ŷ) - 0 2 x̂
4 d
2 d
4 d

 I
  0 (I1  2I 2 )x̂  0 1 ŷ
4 d
4 d

 0 [-(I1  2I 2 )x̂  I 1 ŷ]
4 d
 -
Magnitude 
0
0
2
2
2
(I1  2I 2 ) 2  I 1 or
2I 1  4I 1 I 2  4I 2
4 d
4 d

I1
Direction  180 o  tan 1 
 I 1  2I 2

 counterclockwise from the  x axis.

12. The two wires shown in the figure below are separated by d and carry currents of I in
opposite directions.
(a) Find the magnitude and direction of the net magnetic field at a point midway between the
wires. (b) Find the magnitude and direction of the net magnetic field at point P1, d to the
right of the wire on the right. (c) Find the magnitude and direction of the net magnetic
field at point P2, 2d to the left of the wire on the left.
Solution:
(a)
I2
I1
B1 B2


As shown in the figure, both B1 and B 2 are pointing into the page.

 I
0I
 I

 0
B1  0
d
d
2 r1
2 ( )
2

 I
0I
 I

 0
B2  0
d
d
2 r2
2 ( )
2
 

 I
 I
 B  B1  B 2  0
 0
d
d
2 0 I

d
I2
I1
(b)
B1

I
0I
I

 0
B1  0
4 d
2 (2d)
2 r1

I
0 I

B2  0
2 d
2 r2
 

I
I
 0
 B  B1  B 2  - 0
2 d
4 d
I
 0
4 d
(c)
I1
B1
-
0 I
4 d
I2
B2

I
0 I
I

 0
B1  0
4 d
2 (2d)
2 r1

I
0I
0 I


B2  0
6d
2 (3d)
2 r2
 

I
I
 0
 B  B1  B 2  0
6 d
4 d
0I

12 d
-
0 I
6d
B2