Download B-field hw3 ANS

1.
1A
(a) The force acting on AB points downwards.
1A
The force acting on CD points upwards.
(b) This is because the force acting on BC is parallel to the rotation axis.
1A
(c)
1A
The coil rotates in anticlockwise direction.
(d) X is a commutator.
1A
It changes contact from one brush to the other every half turn of the
coil.
1A
This reverses the current through the coil,
so do the forces acting on the coil.
1A
The coil, therefore, carries on rotating in the same direction.
3  1A
(e) Any three of the following:
Use stronger magnets.
Increase the number of turns in the coil.
Increase the area of the coil within the magnetic field.
Increase the current.
2
(i)
N
(ii)
S
(Correct shape of magnetic field lines)
1A
(Correct direction of magnetic field lines)
1A
(Correct polarities)
1A
An upward force acts on AB and
1A
a downward force acts on CD.
1A
(iii) The coil rotates in clockwise direction as seen by the observer.
(iv)
It changes contact from one brush to the other every half turn of the coil.
1A
(Cancelled)
This reverses the current through the coil,
so do the forces acting on the coil.
The coil, therefore, carries on rotating in the same direction.
(v)
The rotation direction of the motor remains unchanged.
1A
It is because both the current direction and the polarities of the
solenoids are reversed in reversing the battery.
1A
3
Solutions
(a) Since the electric force balances the magnetic force and E =
Marks
VH
,
d
we have QE = BQv.
v=
1M
1 10 6
E VH
=
=
= 5  10–5 m s–1
B Bd (2)(0.01)
1A
The drift velocity of the charge carriers is 5  10–5 m s–1.
(b) By VH =
n=
BI
,
nQt
1M
(2)(3)
BI
=
= 7.5  1027 m–3
5
VH Qt (5  10 )(1.6  10 19 )(0.005)
1A
There are 1.5  1026 charge carriers per unit volume.
(c)
Since the pointer of the centre-zero galvanometer deflects to the
right, the current flows to the right.
1A
By Fleming’s left-hand rule, the charge carriers are deflected
upwards.
1A
4
Solutions
Marks
(a) Downwards
1A
(b) Upper side
1A
(c)
Since the electric force balances the magnetic force and E =
we have QE = BQv.
v=
4  10 6
E VH
=
=
= 0.392 m s–1
B Bd (1.2  10 3 )(0.85  10 2 )
VH
,
d
1M
1A