Magnetic Effects of Current- 2 Practice Questions - DPP for NEET

 Page 1


(1) (b) E
kp
 = eV , \ E
k
 = qV ,
\ E
k
 µ q, \ V = constant
E
kp
 : E
kd 
: E
ka
 : : 1 : 1 : 2.
(2) (c)E
K
 = 
2 22
q rB
2m
\ E
k
 µ 
2 2
p
a
2
k
p p
m
q q
mE
m
=´
= 
41
14
´ = 
E
K
a
 = 8eV. .
(3) (a) B = 
2
KVe
r
 = 
7 6 19
102
10 3 1.6 10
(210)
--
-
´´´
´
   = 1.2 Tesla.
(4) (a) F = qvB sinq
= 2 × 1.6 × 10
–19
 × 10
5
 × 0.8 × 
1
2
æö
ç÷
èø
= 1.28 × 10
–14
 N [ Q charge on a particle = 2e]
(5) (b) The direction of 
r
F
 is along (V B) ´
rr
 which is towards
the right. Thus the beam deflects to your right side.
(6) (b) The particle is moving clockwise which shows that
force on the particle is opposite to given by right hand
palm rule of fleming left hand rule. These two laws are
used for positive charge. Here since laws are disobeyed,
we can say that charge is negative.
(7) (b) The point lies at the circumference hence it will come
back after a time period T
T = 
2p m
qB
(8) (a) The magnetic force on a current carrying wire of length
L, placed in a magnetic field B at an angleq with the
field is given by
F = i l Bsinq.
Here B = 5.0 × 10
–4
 N/A.m. i = 2.0 A,
l = 50 cm = 0.50 m,
q = 60º
F = 2.0 × 0.50 × (5.0 × 10
–4
) × sin 60º
= 4.33 × 10
–4
 N
According to the flemings left - hand rule, this force
will act perpendicular to both the wire and the magnetic
field.
(9) (a, c) r µ 
m
q
 Þ r
H
 : r
He
 : r
o
 = 
1416
::
1 12
 = 1 : 2 : 2
Radius is smallest for H
+
, so it is deflected most.
(10) (b) In the figure, the z-axis points out of the paper, and the
magnetic field is directed into the paper, existing in the
region between PQ and RS. The particle moves in a
circular path of radius r in the magnetic field. It can just
enter the region x > b for r ³ (b – a)
Now r = 
( )
B
mv
ba
q
³-
or v ³ 
( ) B qba
m
-
 Þ v
min
 = 
( ) B qba
m
-
(11) (a) From figure it is clear that
q
b r
q
d
q
sin q = 
d
r
 also r = 
B
p
q
\ sin q = 
Bqd
p
(12) (a) For on wire Q due to wire P is
F
P
 =
7
2 30 10
10 0.1
0.1
-
´´
´´
 = 6 ´ 10
– 5
 N (T owards left)
Force on wire Q due to wire R is
F
R
 = 
7
2 20 10
10 0.1
0.02
-
´´
´´
 = 20 ´ 10
–5
 (T owards right)
Hence F
net 
= F
R
 – F
P
 = 14 ´ 10
–5
N = 14 ´ 10
–4
 N
(Towards right)
(13) (d) F q (v B) =´
rr
r
vB ´
r
r
 = 
5
ˆ ˆˆ
i jk
3 20
5 10 00 ´
 = 
ˆ
k (–10 × 10
5
)   = (–
ˆ
k 10
6
)
q = 2e = 2 × 1.6 x 10
-19
 = 3.2 × 10
-19
 Coulomb
F
r
 = 3.2 × 10
–19
 (–
ˆ
k × 10
6
)
Þ F
r
 = – 3.2 × 10
–13
 
ˆ
k .
\ |F| = 3.2 × 10
–13
 Coulomb.
(14) (b)
\ F = q (
v
r
 × 
r
B
) = 2evB sin 90º
or F = 2evB
Page 2


(1) (b) E
kp
 = eV , \ E
k
 = qV ,
\ E
k
 µ q, \ V = constant
E
kp
 : E
kd 
: E
ka
 : : 1 : 1 : 2.
(2) (c)E
K
 = 
2 22
q rB
2m
\ E
k
 µ 
2 2
p
a
2
k
p p
m
q q
mE
m
=´
= 
41
14
´ = 
E
K
a
 = 8eV. .
(3) (a) B = 
2
KVe
r
 = 
7 6 19
102
10 3 1.6 10
(210)
--
-
´´´
´
   = 1.2 Tesla.
(4) (a) F = qvB sinq
= 2 × 1.6 × 10
–19
 × 10
5
 × 0.8 × 
1
2
æö
ç÷
èø
= 1.28 × 10
–14
 N [ Q charge on a particle = 2e]
(5) (b) The direction of 
r
F
 is along (V B) ´
rr
 which is towards
the right. Thus the beam deflects to your right side.
(6) (b) The particle is moving clockwise which shows that
force on the particle is opposite to given by right hand
palm rule of fleming left hand rule. These two laws are
used for positive charge. Here since laws are disobeyed,
we can say that charge is negative.
(7) (b) The point lies at the circumference hence it will come
back after a time period T
T = 
2p m
qB
(8) (a) The magnetic force on a current carrying wire of length
L, placed in a magnetic field B at an angleq with the
field is given by
F = i l Bsinq.
Here B = 5.0 × 10
–4
 N/A.m. i = 2.0 A,
l = 50 cm = 0.50 m,
q = 60º
F = 2.0 × 0.50 × (5.0 × 10
–4
) × sin 60º
= 4.33 × 10
–4
 N
According to the flemings left - hand rule, this force
will act perpendicular to both the wire and the magnetic
field.
(9) (a, c) r µ 
m
q
 Þ r
H
 : r
He
 : r
o
 = 
1416
::
1 12
 = 1 : 2 : 2
Radius is smallest for H
+
, so it is deflected most.
(10) (b) In the figure, the z-axis points out of the paper, and the
magnetic field is directed into the paper, existing in the
region between PQ and RS. The particle moves in a
circular path of radius r in the magnetic field. It can just
enter the region x > b for r ³ (b – a)
Now r = 
( )
B
mv
ba
q
³-
or v ³ 
( ) B qba
m
-
 Þ v
min
 = 
( ) B qba
m
-
(11) (a) From figure it is clear that
q
b r
q
d
q
sin q = 
d
r
 also r = 
B
p
q
\ sin q = 
Bqd
p
(12) (a) For on wire Q due to wire P is
F
P
 =
7
2 30 10
10 0.1
0.1
-
´´
´´
 = 6 ´ 10
– 5
 N (T owards left)
Force on wire Q due to wire R is
F
R
 = 
7
2 20 10
10 0.1
0.02
-
´´
´´
 = 20 ´ 10
–5
 (T owards right)
Hence F
net 
= F
R
 – F
P
 = 14 ´ 10
–5
N = 14 ´ 10
–4
 N
(Towards right)
(13) (d) F q (v B) =´
rr
r
vB ´
r
r
 = 
5
ˆ ˆˆ
i jk
3 20
5 10 00 ´
 = 
ˆ
k (–10 × 10
5
)   = (–
ˆ
k 10
6
)
q = 2e = 2 × 1.6 x 10
-19
 = 3.2 × 10
-19
 Coulomb
F
r
 = 3.2 × 10
–19
 (–
ˆ
k × 10
6
)
Þ F
r
 = – 3.2 × 10
–13
 
ˆ
k .
\ |F| = 3.2 × 10
–13
 Coulomb.
(14) (b)
\ F = q (
v
r
 × 
r
B
) = 2evB sin 90º
or F = 2evB
DPP/ P 40
114
(15) (a) Force on side BC AND AD are equal but opposite so
their net will be zero.
10 cm
15 cm
2 cm
2 A
A
B C
D
F
AB F
CD
But F
AB
 = 
72
2
2 21
10 15 10
2 10
--
-
´´
´ ´´
´
 = 3 ´ 10
–6
 N
and F
CD
 = 
( )
72
2
2 21
10 15 10
2 10
--
-
´´
´ ´´
´
 = 0.5 ´ 10
–6
 N
Þ F
net
 = F
AB
 – F
CD
 = 2.5 ´ 10
–6
 N
= 25 ´ 10
–7
 N, towards the wire.
(16) (b) In order to make a proton circulate the earth along the
equator, the minimum magnetic field induction 
B
r
should be horizontal nad perpendicular to equator.
The magnetic force provides the necessary centripetal
force.
i.e. qv B = 
2
mv
r
or B = 
mv
qr
Here m = 1.7 × 10
–27
 kg, v = 1.0 × 10
7
 m/s
q = e = 1.6 × 10
-19
 coulomb, r = 6.37 × 10
6
 m
B = 
277
196
1.7 10 1.0 10
1.6 10 6.37 10
-
-
´ ´´
´ ´´
 = 1.67 × 10
–8
 weber/m
2
.
(17) (c) We have F = qvB = 
2
mv
r
 or v = 
qBr
m
= 
19
27
3.2 10 1.2 0.45
6.8 10
-
-
´ ´´
´
 = 2.6 × 10
7
 m/s.
The frequency of rotation n = 
v
2r p
= 
7
2.6 10
2 3.14 0.45
´
´´
 = 9.2 × 10
6
  sec
–1
.
Kinetic energy of a-particle,
E
K
 = 
1
2
 × 6.8 × 10
–27
 × (2.6 × 10
7
)
2
= 2.3 × 10
–12
 joule.
= 
12
19
2.3 10
1.6 10
-
-
´
´
 eV olt = 14 × 10
6
 eV = 14 MeV olt.
If V is accelerating potential of a-particle, then Kinetic
energy = qV
14 × 10
6
 eV olt = 2eV (since charge on a-particle = 2e)
\ V = 
6
14 10
2
´
 = 7 × 10
6
 V olt.
(18) (a) If electron beam passes undeflected in simultaneous
electric and magnetic fields 
r
E
 and 
r
B
 velocity of beam
r
v
 much be mutually perpendicular and the required
speed v is given by-
4
3
E 1 10
v
B
2 10
-
´
==
´
 = 5 × 10
6
 m/s.
If electric field is removed, the electron traverses a
circular path of radius r given by 
2
mv
r
 = evB or r = 
mv
eB
.
Here m = 9.1 × 10
–31
 kg, v = 5 × 10
6
 m/s.
e = 1.6 × 10
–19
 coul and B = 2 × 10
–3
 weber/m
2
\ r = 
316
193
(9.1 10 ) (5 10 )
(1.6 10 ) (2 10 )
-
--
´´
´´
        = 1.43 × 10
-2
 m = 1.43 cm.
(19) (a) For L length or wire, to balance,
F
magnetic
 = mg Þ ILB = mg,
Therefore B = mg/IL = (m/L)g/I
= 
3
45 10 9.8
30
-
´´
 = 1.47 × 10
-2
 tesla.
= 147 Gauss.
(20) (b) According to Fleming's left hand rule, magnetic force
on electrons will be downward.
e
–
e
–
e
–
e
–
e
–
e
–
e
–
×
e
–
e
–
e
–
e
–
e
–
e
–
e
–
×
(21) (b) \ F = mg = Bil
or 1 × 9.8 = 0.98 × i × 1, Þ i = 10A.
(22) (b) When currents flow in two long, parallel wires in the
same direction, the wires exert a force of attraction on
each other. The magnitude of this force acting per meter
length of the wires is given by
F = 
0 12
µ ii
2R p
 = 2 × 10
–7
 
12
ii
R
 N/m.
Here i
1
 = 10 A, i
2
 = 15 A, R = 30 cm = 0.3 m
\ F = 2 × 10
–7
 
10 15
0.3
´
 = 1 × 10
–4
 N/m.
\ Force on 5m length of the wire
= 5 × (1 × 10
–4
) = (5 × 10
-4
) = 5 × 10
-4
 N (attraction).
(23) (d) The electron will pass undeviated if the electric force
and magnetic force are equal and opposite. Thus
E.e. = Bev or B = E/v but E = V/d
Therefore, 
36
V 600
B
v.d.
3 10 2 10
-
==
´ ´´
\ B = 0.1 Wb/m
2
.
The direction of field is perpendicular to the plane of
paper vertically downward.
Page 3


(1) (b) E
kp
 = eV , \ E
k
 = qV ,
\ E
k
 µ q, \ V = constant
E
kp
 : E
kd 
: E
ka
 : : 1 : 1 : 2.
(2) (c)E
K
 = 
2 22
q rB
2m
\ E
k
 µ 
2 2
p
a
2
k
p p
m
q q
mE
m
=´
= 
41
14
´ = 
E
K
a
 = 8eV. .
(3) (a) B = 
2
KVe
r
 = 
7 6 19
102
10 3 1.6 10
(210)
--
-
´´´
´
   = 1.2 Tesla.
(4) (a) F = qvB sinq
= 2 × 1.6 × 10
–19
 × 10
5
 × 0.8 × 
1
2
æö
ç÷
èø
= 1.28 × 10
–14
 N [ Q charge on a particle = 2e]
(5) (b) The direction of 
r
F
 is along (V B) ´
rr
 which is towards
the right. Thus the beam deflects to your right side.
(6) (b) The particle is moving clockwise which shows that
force on the particle is opposite to given by right hand
palm rule of fleming left hand rule. These two laws are
used for positive charge. Here since laws are disobeyed,
we can say that charge is negative.
(7) (b) The point lies at the circumference hence it will come
back after a time period T
T = 
2p m
qB
(8) (a) The magnetic force on a current carrying wire of length
L, placed in a magnetic field B at an angleq with the
field is given by
F = i l Bsinq.
Here B = 5.0 × 10
–4
 N/A.m. i = 2.0 A,
l = 50 cm = 0.50 m,
q = 60º
F = 2.0 × 0.50 × (5.0 × 10
–4
) × sin 60º
= 4.33 × 10
–4
 N
According to the flemings left - hand rule, this force
will act perpendicular to both the wire and the magnetic
field.
(9) (a, c) r µ 
m
q
 Þ r
H
 : r
He
 : r
o
 = 
1416
::
1 12
 = 1 : 2 : 2
Radius is smallest for H
+
, so it is deflected most.
(10) (b) In the figure, the z-axis points out of the paper, and the
magnetic field is directed into the paper, existing in the
region between PQ and RS. The particle moves in a
circular path of radius r in the magnetic field. It can just
enter the region x > b for r ³ (b – a)
Now r = 
( )
B
mv
ba
q
³-
or v ³ 
( ) B qba
m
-
 Þ v
min
 = 
( ) B qba
m
-
(11) (a) From figure it is clear that
q
b r
q
d
q
sin q = 
d
r
 also r = 
B
p
q
\ sin q = 
Bqd
p
(12) (a) For on wire Q due to wire P is
F
P
 =
7
2 30 10
10 0.1
0.1
-
´´
´´
 = 6 ´ 10
– 5
 N (T owards left)
Force on wire Q due to wire R is
F
R
 = 
7
2 20 10
10 0.1
0.02
-
´´
´´
 = 20 ´ 10
–5
 (T owards right)
Hence F
net 
= F
R
 – F
P
 = 14 ´ 10
–5
N = 14 ´ 10
–4
 N
(Towards right)
(13) (d) F q (v B) =´
rr
r
vB ´
r
r
 = 
5
ˆ ˆˆ
i jk
3 20
5 10 00 ´
 = 
ˆ
k (–10 × 10
5
)   = (–
ˆ
k 10
6
)
q = 2e = 2 × 1.6 x 10
-19
 = 3.2 × 10
-19
 Coulomb
F
r
 = 3.2 × 10
–19
 (–
ˆ
k × 10
6
)
Þ F
r
 = – 3.2 × 10
–13
 
ˆ
k .
\ |F| = 3.2 × 10
–13
 Coulomb.
(14) (b)
\ F = q (
v
r
 × 
r
B
) = 2evB sin 90º
or F = 2evB
DPP/ P 40
114
(15) (a) Force on side BC AND AD are equal but opposite so
their net will be zero.
10 cm
15 cm
2 cm
2 A
A
B C
D
F
AB F
CD
But F
AB
 = 
72
2
2 21
10 15 10
2 10
--
-
´´
´ ´´
´
 = 3 ´ 10
–6
 N
and F
CD
 = 
( )
72
2
2 21
10 15 10
2 10
--
-
´´
´ ´´
´
 = 0.5 ´ 10
–6
 N
Þ F
net
 = F
AB
 – F
CD
 = 2.5 ´ 10
–6
 N
= 25 ´ 10
–7
 N, towards the wire.
(16) (b) In order to make a proton circulate the earth along the
equator, the minimum magnetic field induction 
B
r
should be horizontal nad perpendicular to equator.
The magnetic force provides the necessary centripetal
force.
i.e. qv B = 
2
mv
r
or B = 
mv
qr
Here m = 1.7 × 10
–27
 kg, v = 1.0 × 10
7
 m/s
q = e = 1.6 × 10
-19
 coulomb, r = 6.37 × 10
6
 m
B = 
277
196
1.7 10 1.0 10
1.6 10 6.37 10
-
-
´ ´´
´ ´´
 = 1.67 × 10
–8
 weber/m
2
.
(17) (c) We have F = qvB = 
2
mv
r
 or v = 
qBr
m
= 
19
27
3.2 10 1.2 0.45
6.8 10
-
-
´ ´´
´
 = 2.6 × 10
7
 m/s.
The frequency of rotation n = 
v
2r p
= 
7
2.6 10
2 3.14 0.45
´
´´
 = 9.2 × 10
6
  sec
–1
.
Kinetic energy of a-particle,
E
K
 = 
1
2
 × 6.8 × 10
–27
 × (2.6 × 10
7
)
2
= 2.3 × 10
–12
 joule.
= 
12
19
2.3 10
1.6 10
-
-
´
´
 eV olt = 14 × 10
6
 eV = 14 MeV olt.
If V is accelerating potential of a-particle, then Kinetic
energy = qV
14 × 10
6
 eV olt = 2eV (since charge on a-particle = 2e)
\ V = 
6
14 10
2
´
 = 7 × 10
6
 V olt.
(18) (a) If electron beam passes undeflected in simultaneous
electric and magnetic fields 
r
E
 and 
r
B
 velocity of beam
r
v
 much be mutually perpendicular and the required
speed v is given by-
4
3
E 1 10
v
B
2 10
-
´
==
´
 = 5 × 10
6
 m/s.
If electric field is removed, the electron traverses a
circular path of radius r given by 
2
mv
r
 = evB or r = 
mv
eB
.
Here m = 9.1 × 10
–31
 kg, v = 5 × 10
6
 m/s.
e = 1.6 × 10
–19
 coul and B = 2 × 10
–3
 weber/m
2
\ r = 
316
193
(9.1 10 ) (5 10 )
(1.6 10 ) (2 10 )
-
--
´´
´´
        = 1.43 × 10
-2
 m = 1.43 cm.
(19) (a) For L length or wire, to balance,
F
magnetic
 = mg Þ ILB = mg,
Therefore B = mg/IL = (m/L)g/I
= 
3
45 10 9.8
30
-
´´
 = 1.47 × 10
-2
 tesla.
= 147 Gauss.
(20) (b) According to Fleming's left hand rule, magnetic force
on electrons will be downward.
e
–
e
–
e
–
e
–
e
–
e
–
e
–
×
e
–
e
–
e
–
e
–
e
–
e
–
e
–
×
(21) (b) \ F = mg = Bil
or 1 × 9.8 = 0.98 × i × 1, Þ i = 10A.
(22) (b) When currents flow in two long, parallel wires in the
same direction, the wires exert a force of attraction on
each other. The magnitude of this force acting per meter
length of the wires is given by
F = 
0 12
µ ii
2R p
 = 2 × 10
–7
 
12
ii
R
 N/m.
Here i
1
 = 10 A, i
2
 = 15 A, R = 30 cm = 0.3 m
\ F = 2 × 10
–7
 
10 15
0.3
´
 = 1 × 10
–4
 N/m.
\ Force on 5m length of the wire
= 5 × (1 × 10
–4
) = (5 × 10
-4
) = 5 × 10
-4
 N (attraction).
(23) (d) The electron will pass undeviated if the electric force
and magnetic force are equal and opposite. Thus
E.e. = Bev or B = E/v but E = V/d
Therefore, 
36
V 600
B
v.d.
3 10 2 10
-
==
´ ´´
\ B = 0.1 Wb/m
2
.
The direction of field is perpendicular to the plane of
paper vertically downward.
DPP/ P 40
115
(24) (b) The component of velocity of the beam of protons,
parallel to the field direction
= v cosq = 4 × 10
5
 × cos 60º = 2 × 10
5
 m/sec.
and the component of velocity of the proton beam at
right angle to the direction of field
= v sinq = 4 × 10
5
 × sin 60º = 2
3 × 10
5
 m/sec.
therefore,the radius of circular path = (mv sinq /Be)
or  r = 
275
19
1.7 10 2 3 10
0.3 1.6 10
-
-
´ ´´
´´
= 12.26 × 10
–3
 metre
or r = 1.226 × 10
–2
 metre.
Pitch of the Helix = v cosq x (2pm/Be)
\ Pitch = 
5 27
19
2 10 2 3.14 1.7 10
0.3 1.6 10
-
-
´ ´´ ´´
´´
        = 44.5 × 10
–3
 m = 4.45 × 10
–2
 m.
(25) (a), (26) (a)
F q (v B) =´
rr
r
 = 
22
ˆ
q (x y )k -