Magnetic Effects of Current- 1 Practice Questions - DPP for NEET

 Page 1


(1) (a) We know magnetic field due to a long straight current
carrying wire
B = 
7
0
3
µi 4 103
2r
2 50 10
-
-
p´´
=
p
p´´
(Note that  m
0
 = 4p × 10
–7 
in SI system)
   = 1.20 × 10
–5
 Tesla = 0.12 G .
[As 1 Gauss = 10
–4
 Tesla]
(2) (c) The magnetic induction produced due to a current
carrying arc at its centre of curvature is
B = 
0
i
4r
ma
p
.......... (a)
(subtending angle a at the centre of curvature)
Þ
00
µii
B
4 r 4 16r
pm p
= ´=
p
(3) (a)
90°
O
Q
R
r
i
P
T
x
y
z
i
S
O QRS ST
B B B. =+
ur ur u r
00
PQ QRS ST
µ i µi 3
ˆˆ
B zero, B k, B k
4 2r 4r
= =´=
p
ur u r ur
Þ
0 00
O
µi 3µi µi 33
ˆ ˆˆ
B k k 1k
4r 4 2r 4r 2
p éù
= + =+
êú
ëû pp
ur
(4) (a).
O PSR PQR
B BB =+
ur ur u r
.... (a)
PSR B
ur
 = 
00
ii 22
4 r 2r
mm p-f éù
=
êú
pp
ëû
 [p - f]
ˆ
( k) - .... (b)
PQR
B
ur
 
0 00
i ii 2sin 2sin
ˆˆ
. (k) . tan (k)
4 OQ 4 rcos 2r
m mm ff
= - = = f-
p p fp
.... (c)
From eqs. (a), (b) and (c)
   B
ur
 = 
0
i
2r
m
p
 [p - f]
ˆ
( k) - + 
0
i
2r
m
p
 tan f 
ˆ
( k) -
=  
0
i
2r
m
p
 [p – f + tan f]
ˆ
( k) -
(5) (a). The rotating rod is a current-loop whose radius a = 0.6m.
The magnetic field due to this current-loop at a point
on its axis at a distance x from its centre is given by
2
0
2 2 3/2
µ ia
B
2(a x)
=
+
...(i)
Let T be the period of rotation of the rod. Then
4
q q 1 colulomb 10 / sec
i
T22
w ´p
= ==
pp
    = 5 × 10
3
 amp.
Now, a = 0.6 m, x = 0.8 m and m
0
 = 4p x 10
–7
 V -s/A-m.
Substituting these values in eq. (i) we get
7 32
3/23
(4 10 V s / A m)(5 10 A)(0.6m)
B
2(0.36 0.64) m
-
p´ - -´
=
+
    = 0.36p × 10
–3 
 = 1.13 × 10
–3
 tesla
In the second case the current remains the same because
the rotating charge and the angular frequency are the
same. However, the radius of the loop becomes half (a
= 0.3 m) and the distance x is now 0.4 m.
\ 
2
0
2 2 3/2
µ ia
B
2(a x)
=
+
= 
7 32
3/23
(4 10 V s / A m)(5 10 A)(0.3m)
2(0.09 0.16) m
-
p´ - -´
+
= 
73
4 10 5 10 0.09
2 0.25 0.5
-
p´ ´´´
´´
 tesla
= 0.72 × 10
–3
 p = 2.26 × 10
–3
 tesla.
(6) (a) The magnetic field at the centre of a current carrying
coil having n turns is given by
B = 
m
0
2
ni
r
 N/A.m
where i, is the current in the coil and r is the radius of
the coil.
Here i = 0.1 A, n = 1000 and r = 0.1 m.
\ B = 
7
(4 10 ) 1000 0.1
2 0.1
-
p´ ´´
´
 = 6.28 × 10
-4
 N/A.m
(7) (a). The two coils are perpendicular to each other. Coil 1
produces field along X axis and coil 2 produces field
along Y axis. Thus the resultant field will be-
B = B B
1
2
2
2
+ making an angle
1 2
1
B
tan
B
-
æö
q=
ç÷
èø
 with x axis
AsB
1
 = B
2
 
0
µ NI
2a
=
Þ B = 
2
 
00
µ NI µ NI
2a 2a
æö
==
ç÷
èø
 and  q = 45º.
Page 2


(1) (a) We know magnetic field due to a long straight current
carrying wire
B = 
7
0
3
µi 4 103
2r
2 50 10
-
-
p´´
=
p
p´´
(Note that  m
0
 = 4p × 10
–7 
in SI system)
   = 1.20 × 10
–5
 Tesla = 0.12 G .
[As 1 Gauss = 10
–4
 Tesla]
(2) (c) The magnetic induction produced due to a current
carrying arc at its centre of curvature is
B = 
0
i
4r
ma
p
.......... (a)
(subtending angle a at the centre of curvature)
Þ
00
µii
B
4 r 4 16r
pm p
= ´=
p
(3) (a)
90°
O
Q
R
r
i
P
T
x
y
z
i
S
O QRS ST
B B B. =+
ur ur u r
00
PQ QRS ST
µ i µi 3
ˆˆ
B zero, B k, B k
4 2r 4r
= =´=
p
ur u r ur
Þ
0 00
O
µi 3µi µi 33
ˆ ˆˆ
B k k 1k
4r 4 2r 4r 2
p éù
= + =+
êú
ëû pp
ur
(4) (a).
O PSR PQR
B BB =+
ur ur u r
.... (a)
PSR B
ur
 = 
00
ii 22
4 r 2r
mm p-f éù
=
êú
pp
ëû
 [p - f]
ˆ
( k) - .... (b)
PQR
B
ur
 
0 00
i ii 2sin 2sin
ˆˆ
. (k) . tan (k)
4 OQ 4 rcos 2r
m mm ff
= - = = f-
p p fp
.... (c)
From eqs. (a), (b) and (c)
   B
ur
 = 
0
i
2r
m
p
 [p - f]
ˆ
( k) - + 
0
i
2r
m
p
 tan f 
ˆ
( k) -
=  
0
i
2r
m
p
 [p – f + tan f]
ˆ
( k) -
(5) (a). The rotating rod is a current-loop whose radius a = 0.6m.
The magnetic field due to this current-loop at a point
on its axis at a distance x from its centre is given by
2
0
2 2 3/2
µ ia
B
2(a x)
=
+
...(i)
Let T be the period of rotation of the rod. Then
4
q q 1 colulomb 10 / sec
i
T22
w ´p
= ==
pp
    = 5 × 10
3
 amp.
Now, a = 0.6 m, x = 0.8 m and m
0
 = 4p x 10
–7
 V -s/A-m.
Substituting these values in eq. (i) we get
7 32
3/23
(4 10 V s / A m)(5 10 A)(0.6m)
B
2(0.36 0.64) m
-
p´ - -´
=
+
    = 0.36p × 10
–3 
 = 1.13 × 10
–3
 tesla
In the second case the current remains the same because
the rotating charge and the angular frequency are the
same. However, the radius of the loop becomes half (a
= 0.3 m) and the distance x is now 0.4 m.
\ 
2
0
2 2 3/2
µ ia
B
2(a x)
=
+
= 
7 32
3/23
(4 10 V s / A m)(5 10 A)(0.3m)
2(0.09 0.16) m
-
p´ - -´
+
= 
73
4 10 5 10 0.09
2 0.25 0.5
-
p´ ´´´
´´
 tesla
= 0.72 × 10
–3
 p = 2.26 × 10
–3
 tesla.
(6) (a) The magnetic field at the centre of a current carrying
coil having n turns is given by
B = 
m
0
2
ni
r
 N/A.m
where i, is the current in the coil and r is the radius of
the coil.
Here i = 0.1 A, n = 1000 and r = 0.1 m.
\ B = 
7
(4 10 ) 1000 0.1
2 0.1
-
p´ ´´
´
 = 6.28 × 10
-4
 N/A.m
(7) (a). The two coils are perpendicular to each other. Coil 1
produces field along X axis and coil 2 produces field
along Y axis. Thus the resultant field will be-
B = B B
1
2
2
2
+ making an angle
1 2
1
B
tan
B
-
æö
q=
ç÷
èø
 with x axis
AsB
1
 = B
2
 
0
µ NI
2a
=
Þ B = 
2
 
00
µ NI µ NI
2a 2a
æö
==
ç÷
èø
 and  q = 45º.
DPP/ P 39
111
(8) (d) Applying ampere's law at P , Q and R respectively , we
find that there is no current enclosed by the circle of P .
So magnetic induction at P is zero while that at Q and
R is non- zero.
(9) (a). For a current carrying coil
B = 
0
µi
2R
 at centre and force on a current carrying
conductor ie
 F = i l B Þ F = 
2
0
µi
2R
l
Þ [MLT
–2
] = 
2
0
[µ ][A ][L]
[L]
Þ [m
0
] = [MLT
–2
 A
–2
]
(10) (c) By Biat Savart Law,
dB = 
0
2
i sin
4
r
m dq
p
l
When q = 90º, then sin 90º = 1 = maximum
\ dB = 
0
2
i
4r
md
p
l
 = maximum
(11) (a) The magnitude of the magnetic field at the centroid O
of the triangle due to a side PQ (say) is
0
12
µ i
(sin sin)
4r
f+f
p
Where r is the perpendicular distance of PQ from O,
and f
1
, f
2 
 the angles as shown. The field is
perpendicular to the plane of paper and is directed into
plane of paper. Since the magnetic field due to each of
the three sides is the same in magnitude and direction,
the magnitude of the resultant field at O is
Q
P R
O
(
f
2
(
f
1
i
B = 
0
12
µ i
3 (sin sin )
4r
f+f
p
Here i = 1 ampere, f
1
 =  f
2
 = 60º
and  r = 
2
l
 cot 60º = 
1
2 3
´
l
and l is the side of the triangle (= 4.5 × 10
–2
 meter).
 \ B = 
7
2
3 10 1.0
11
4.5 10
2 3
-
-
´´
æö æö
´ ´´
ç÷ç÷
èøèø
33
22
æö
+
ç÷
ç÷
èø
         =
7
2
3 10 23
4.5 10
-
-
´ ´´
´
 = 4.0 × 10
–5
 weber/m
2
.
(12) (a). The magnetic field inside (near centre) a current carrying
solenoid having n turns per unit length is given by
B = m
0
 ni newton/(ampere-meter),
where i (ampere) is the current in the solenoid and
m
0
 = 4p × 10
–7
 newton/ampere
2
.
Here n = 500/0.40 = 1250 per meter, i = 1.0 amp.
\ B = (4 × 3.14 × 10
-7
) ×  1250 × 1.0
       = 15.7 × 10
-4
 newton/(ampere-meter) = 15.7 gauss.
(13) (b) We know , B
end
 = 
0
ni
2
m
Here n = 
500
0.2
 = 2500/metre,
\ i = 
6
end
7
0
2B 2 8.71 10
µn
4 10 2500
-
-
´´
=
p´´
  = 
3
17.42 10 0.01742
amp
-
´
=
pp
 amp.
(14) (b)
centre
axis
B
B
 = 
3/2
2
2
1
R
x
æö
´ç÷
ç÷
èø
, also Baxis = centre
1
B
8
Þ
8
1
=  
3/2
2
2
1
R
x
æö
´ç÷
ç÷
èø
 Þ 2 = 
1/2
2
2
1
R
x
æö
´ç÷
ç÷
èø
Þ 4 = 1 + 
2
2
R
x
 Þ 3 = 
2
2
R
x
 Þ x
2
 = 3R
2
 Þ 3R
2
Þ x = 3R
(15) (a) B
0
 = m
0
 
Ni
2R p
= 
7
4 10 500 0.5
2 0.1
-
p´ ´´
p´
 = 5 × 10
–4
 tesla
(16) (c) At P : B
net
 = 
22
12
BB +
= 
22
00 12
µµ 22
4p 4p
ii
aa
æ öæö
+
ç ÷ç÷
è øèø
A
B
C D
i
1
i
2
B
1
B
2
P
O
a
= ( )
1/2
22 0
12
µ
2p
ii
a
+
Page 3


(1) (a) We know magnetic field due to a long straight current
carrying wire
B = 
7
0
3
µi 4 103
2r
2 50 10
-
-
p´´
=
p
p´´
(Note that  m
0
 = 4p × 10
–7 
in SI system)
   = 1.20 × 10
–5
 Tesla = 0.12 G .
[As 1 Gauss = 10
–4
 Tesla]
(2) (c) The magnetic induction produced due to a current
carrying arc at its centre of curvature is
B = 
0
i
4r
ma
p
.......... (a)
(subtending angle a at the centre of curvature)
Þ
00
µii
B
4 r 4 16r
pm p
= ´=
p
(3) (a)
90°
O
Q
R
r
i
P
T
x
y
z
i
S
O QRS ST
B B B. =+
ur ur u r
00
PQ QRS ST
µ i µi 3
ˆˆ
B zero, B k, B k
4 2r 4r
= =´=
p
ur u r ur
Þ
0 00
O
µi 3µi µi 33
ˆ ˆˆ
B k k 1k
4r 4 2r 4r 2
p éù
= + =+
êú
ëû pp
ur
(4) (a).
O PSR PQR
B BB =+
ur ur u r
.... (a)
PSR B
ur
 = 
00
ii 22
4 r 2r
mm p-f éù
=
êú
pp
ëû
 [p - f]
ˆ
( k) - .... (b)
PQR
B
ur
 
0 00
i ii 2sin 2sin
ˆˆ
. (k) . tan (k)
4 OQ 4 rcos 2r
m mm ff
= - = = f-
p p fp
.... (c)
From eqs. (a), (b) and (c)
   B
ur
 = 
0
i
2r
m
p
 [p - f]
ˆ
( k) - + 
0
i
2r
m
p
 tan f 
ˆ
( k) -
=  
0
i
2r
m
p
 [p – f + tan f]
ˆ
( k) -
(5) (a). The rotating rod is a current-loop whose radius a = 0.6m.
The magnetic field due to this current-loop at a point
on its axis at a distance x from its centre is given by
2
0
2 2 3/2
µ ia
B
2(a x)
=
+
...(i)
Let T be the period of rotation of the rod. Then
4
q q 1 colulomb 10 / sec
i
T22
w ´p
= ==
pp
    = 5 × 10
3
 amp.
Now, a = 0.6 m, x = 0.8 m and m
0
 = 4p x 10
–7
 V -s/A-m.
Substituting these values in eq. (i) we get
7 32
3/23
(4 10 V s / A m)(5 10 A)(0.6m)
B
2(0.36 0.64) m
-
p´ - -´
=
+
    = 0.36p × 10
–3 
 = 1.13 × 10
–3
 tesla
In the second case the current remains the same because
the rotating charge and the angular frequency are the
same. However, the radius of the loop becomes half (a
= 0.3 m) and the distance x is now 0.4 m.
\ 
2
0
2 2 3/2
µ ia
B
2(a x)
=
+
= 
7 32
3/23
(4 10 V s / A m)(5 10 A)(0.3m)
2(0.09 0.16) m
-
p´ - -´
+
= 
73
4 10 5 10 0.09
2 0.25 0.5
-
p´ ´´´
´´
 tesla
= 0.72 × 10
–3
 p = 2.26 × 10
–3
 tesla.
(6) (a) The magnetic field at the centre of a current carrying
coil having n turns is given by
B = 
m
0
2
ni
r
 N/A.m
where i, is the current in the coil and r is the radius of
the coil.
Here i = 0.1 A, n = 1000 and r = 0.1 m.
\ B = 
7
(4 10 ) 1000 0.1
2 0.1
-
p´ ´´
´
 = 6.28 × 10
-4
 N/A.m
(7) (a). The two coils are perpendicular to each other. Coil 1
produces field along X axis and coil 2 produces field
along Y axis. Thus the resultant field will be-
B = B B
1
2
2
2
+ making an angle
1 2
1
B
tan
B
-
æö
q=
ç÷
èø
 with x axis
AsB
1
 = B
2
 
0
µ NI
2a
=
Þ B = 
2
 
00
µ NI µ NI
2a 2a
æö
==
ç÷
èø
 and  q = 45º.
DPP/ P 39
111
(8) (d) Applying ampere's law at P , Q and R respectively , we
find that there is no current enclosed by the circle of P .
So magnetic induction at P is zero while that at Q and
R is non- zero.
(9) (a). For a current carrying coil
B = 
0
µi
2R
 at centre and force on a current carrying
conductor ie
 F = i l B Þ F = 
2
0
µi
2R
l
Þ [MLT
–2
] = 
2
0
[µ ][A ][L]
[L]
Þ [m
0
] = [MLT
–2
 A
–2
]
(10) (c) By Biat Savart Law,
dB = 
0
2
i sin
4
r
m dq
p
l
When q = 90º, then sin 90º = 1 = maximum
\ dB = 
0
2
i
4r
md
p
l
 = maximum
(11) (a) The magnitude of the magnetic field at the centroid O
of the triangle due to a side PQ (say) is
0
12
µ i
(sin sin)
4r
f+f
p
Where r is the perpendicular distance of PQ from O,
and f
1
, f
2 
 the angles as shown. The field is
perpendicular to the plane of paper and is directed into
plane of paper. Since the magnetic field due to each of
the three sides is the same in magnitude and direction,
the magnitude of the resultant field at O is
Q
P R
O
(
f
2
(
f
1
i
B = 
0
12
µ i
3 (sin sin )
4r
f+f
p
Here i = 1 ampere, f
1
 =  f
2
 = 60º
and  r = 
2
l
 cot 60º = 
1
2 3
´
l
and l is the side of the triangle (= 4.5 × 10
–2
 meter).
 \ B = 
7
2
3 10 1.0
11
4.5 10
2 3
-
-
´´
æö æö
´ ´´
ç÷ç÷
èøèø
33
22
æö
+
ç÷
ç÷
èø
         =
7
2
3 10 23
4.5 10
-
-
´ ´´
´
 = 4.0 × 10
–5
 weber/m
2
.
(12) (a). The magnetic field inside (near centre) a current carrying
solenoid having n turns per unit length is given by
B = m
0
 ni newton/(ampere-meter),
where i (ampere) is the current in the solenoid and
m
0
 = 4p × 10
–7
 newton/ampere
2
.
Here n = 500/0.40 = 1250 per meter, i = 1.0 amp.
\ B = (4 × 3.14 × 10
-7
) ×  1250 × 1.0
       = 15.7 × 10
-4
 newton/(ampere-meter) = 15.7 gauss.
(13) (b) We know , B
end
 = 
0
ni
2
m
Here n = 
500
0.2
 = 2500/metre,
\ i = 
6
end
7
0
2B 2 8.71 10
µn
4 10 2500
-
-
´´
=
p´´
  = 
3
17.42 10 0.01742
amp
-
´
=
pp
 amp.
(14) (b)
centre
axis
B
B
 = 
3/2
2
2
1
R
x
æö
´ç÷
ç÷
èø
, also Baxis = centre
1
B
8
Þ
8
1
=  
3/2
2
2
1
R
x
æö
´ç÷
ç÷
èø
 Þ 2 = 
1/2
2
2
1
R
x
æö
´ç÷
ç÷
èø
Þ 4 = 1 + 
2
2
R
x
 Þ 3 = 
2
2
R
x
 Þ x
2
 = 3R
2
 Þ 3R
2
Þ x = 3R
(15) (a) B
0
 = m
0
 
Ni
2R p
= 
7
4 10 500 0.5
2 0.1
-
p´ ´´
p´
 = 5 × 10
–4
 tesla
(16) (c) At P : B
net
 = 
22
12
BB +
= 
22
00 12
µµ 22
4p 4p
ii
aa
æ öæö
+
ç ÷ç÷
è øèø
A
B
C D
i
1
i
2
B
1
B
2
P
O
a
= ( )
1/2
22 0
12
µ
2p
ii
a
+
DPP/ P 39
112
(17) (d)
(18) (b) Current distribution in the network is as shown.
Now , consider the pair of wires AB and GH. As current
in these wires produce equal but opposite magnetic
fields at centre O of the cube, resultant field due to the
pair is zero.
E
F
C B
D
O
H
I
I
A
G
I
6
–
I
6
–
I
6
–
I
6
–
I
6
–
I
6
–
I
6
–
I
3
–
I
3
–
I
3
–
I
3
–
I
3
–
We can see five such more pairs namely :
(i) AE, CG
(ii) AD, FI
(iiii) B C, EH
(iv) EF , DC
(v) BF, OH
Magnetic field due to each of these pairs is zero.
Therefore, resultant magnetic field at centre O is zero.
(19) (a) Magnetic field inside a solid cylinder of current is
0
inside
2
µ ir
B
2R
=
p
Þ
0
0
2
R
µi
2
B
2R
=
p
(as per given
information)
Þ
0
0
4BR
i
µ
p
=
Magnetic field outside a solid cylinder of current is
0
outside
µi
B
2r
=
p
Þ B
outside
 at a distance  
0
0
0
0
4BR
µ
µ
2RB
2 (2R)
p æö
ç÷
èø
==
p
(20) (d) As per sense of transversal,
i
crossing
 = I
1
 – I
2
 – I
3
By Ampere's law, B.de
ò
r
r
Ñ
 = µ
0
 i
crossing
Þ B.de
ò
r
r
Ñ
 = µ
0
 (I
2
 – I
1
 – I
3
)
(21) (a) l = (2pr) n or n = 
2r p
l
B = 
00
2
nii
2r
4r
mm
=
p
l
or B = 
7
2
4 10 6.28 1
2 2 (0.10)
-
p´ ´´
´ ´p´
 = 6.28 × 10
-5
 Tesla.
(22) (b). The arrangement is shown in fig.
I
2
I
1
X
R
P
Y
A
The magnetic field at a point P in between the two
wires is
12
B BB =+
r rr
. The field B
1
 (due to current I
1
) points
down ward while B
2
 (due to current I
2
) points upwards.
Thus field at point P is-
B = 
012
II
2 x Rx
méù
-
êú
p-
ëû
 in to the plane of paper. .
At x = R/2,
B = 
0 12
(I I)
R
m-
p
 into the plane of paper,  (if I
1
 > I
2
)
or B = 
0 21
(I I)
R
m-
p
 out of the plane of paper (if I
2
 > I
1
)
(23) (d) (i) Fields due to both coils are in the same direction
Þ B = 
0 11 0 22
12
µNI µ NI
2R 2R
+
IfI
1
 = I
2
 = I, N
1
 = N
2
 = N,
B = 
0 12
12
µ NI(R R)
2RR
+
(ii) Fields due to the two coils are in opposite direction,
Þ B = 
0 11 0 22
12
µNI µNI
2R 2R
-
IfI
1
 = I
2
 = I, N
1
 = N
2
 = N,
B = 
0 21
12
µ NI(R R)
2RR
-
(24) (a). For circular coil B
1
 = 
0
I
2r
m
Circumference of the coil = 2pr = L.
Thus B
1
 = pm
0
 I/L = 3.14 m
0
 I/L
For square loop B
2
 = 2 2 m
0
 I/L = 3.60 m
0
 I/L
Thus B
1
 < B
2
.
25. (d) Since MB
rr
P \ Torque MB =´
uu r ur
 is zero.
26. (d) The field must be in + 
$
k direction.
27. (a)
28. (b) The statements are independently correct.
29. (d)
mB0 t= ´ Þ t=
r uur urr
 for q = 0°, 180°.
30. (b)