EMF Exercise Class 7 Answers
Q1: (a) For the ions to be undeflected in the velocity selector, the electric and magnetic forces must cancel out:
Magnitude of electric force: F
E
= eE
Magnitude of magnetic force: F
M
= evB
1
So v = E/B
1
(proved in lectures).
F
M
= e( v x B
1
) v
Therefore the selected ions all enter the region of field B
2
with speed v = E/B
1
.
They now experience a magnetic force F
M
 e ( v

B
2
)
Magnitude : evB
2
Direction : Perpendicular to v and B
2
As shown in the lectures, the ions will therefore describe a circular path
Let r = radius of circle m = mass of ion
Force on ion is F
M
= evB
2
= mv 2 /r
 r
 mv eB
2
 mE eB
1
B
2
Impact distance L = 2r, so L

2 mE eB
1
B
2
Therefore m
 eB
1
B
2 E
2
L
(b) For two ions of masses m
1
and m
2
, we have
L
L
1
2
 m m
1
2
.
.
F
E
= e E
Therefore m
2



38 .
26
29 .
20


16 = 18.0 units.

Q2 Consider one of the loops as seen from above. The contribution of a small element, d L , to the magnetic field at the centre, O, is given by d B


The magnitude of
0
4
 r
I
2
( d L d L x r
ˆ x
ˆ
)
is just dL because are perpendicular, and the magnitude of d L and r
ˆ
is 1. r
ˆ
O d L
I
By the right hand rule, the direction of d L x r is perpendicular to the coil, pointing into the page if the current is clockwise as shown here, or out of the page if it is anticlockwise.
Let k
ˆ
be a unit vector pointing out of the page.
Then d B
 

4
0

IdL ˆ r
2
depending on the direction of the current.
To find the total field, we integrate one quarter of the way around a circular loop of radius r i.e., from L = 0 to L = (1/4)(2
 r) =
 r/2, to get
B
 



0
4
 r
I
2


2 L
  r
L


0
/ dL
ˆ  

0
8 r
I ˆ
.
The straight sections don’t contribute to B because their d L and r
ˆ
vectors are parallel.
The field at O will have therefore have two contributions, one from the inner loop and one from the outer loop.
Inner loop: B inner

 o
8 a
I ˆ
Therefore B = B inner
+ B outer

 o
8
I

1 a
Outer loop:
1 b


ˆ
B outer

-

B

 o
8 b
I ˆ
 o
I ( b
8 ab
 a ) ˆ .