Magnetic Effects of Current- 3 Practice Questions - DPP for NEET

 Page 1


1. ( b)
222
22
20 (4 10 ) 3 0.3
7
M NiA Am
-
= = ´ ´ ´=-
2. (a) The magnetic moment of current carrying loop
M = niA = ni (pr
2
)
Hence the work done in rotating it through 180°
W = MB(1 – cosq) = 2MB = 2(nipr
2
)B
= 2 × (50 × 2 × 3.14 ×16 × 10
–4
) × 0.1 = 0.1J
3. ( b)
(Number of turns)
NiAB
N
C
q = Þ qµ
4. ( d)
max
sin MB NiAB t= qÞt= , (q = 90°)
5. (c) In equilibrium angle between M
uur
 and B
ur
is zero. It
happens, when plane of the coil is perpendicular to B
ur
B
M
6. (a) t = NBiA = 100 × 0.2 × 2 × (0.08 × 0.1) = 0.32 N × m
7. (c) t = NBiA = 100 × 0.5 × 1 × 400 × 10
–4
 = 2 N–m
8. (a) t = NiAB sinq = 0 ( 0) q=° Q
9. (c) M = NiA Þ M µ A Þ M µ r
2
(As l = 2p r Þ lµ r )
Þ M µ l
2
10. (a)
11. (c)
2
max
1 () NiAB i rB t = =´´p´
2,
2
L
r Lr
æö
p = Þ=
ç÷
èø p
2 2
max
24
L L iB
iB
æö
t=p=
ç÷
èøpp
12. (b)
13. (b) The magnetic field at the centre of the solenoid is
B = m
0
 ni = (4p × 10
-7
) × (500/0.4) × 3 N/A.m.
The torque acting on a current-carrying coil having N
turns (say), placed perpendicular to the axis at the centre
of the solenoid is-
t = BiNA = (4p × 10
-7
) × (500/0.4) × 3 × 0.4 × 10
× p (0.01)
2
 = 6p
2
 × 10
-7 
= 5.92 × 10
-6
 N.m.
14. (d) The equivalent magnetic moment is
M = iA = ef (pr
2
)
As  f = 
v
2r p
\ M = 
ev
2r p
pr
2
 = 
evr
2
15. (b)
C
ii
NAB
q
= Þ µq
16. (d) Initially for circular coil L = 2pr and M = i × pr
2
= 
2 2
24
L iL
i
æö
´p=
ç÷
èøpp
     ...(i)
Finally for square coil
2 2
'
4 16
L iL
Mi
æö
=´=
ç÷
èø
    ...(ii)
L/4
i
Solving equation (i) and (ii) 
4
M
M
p
¢=
17. (b)
2
M iA iR = = ´p
also 
2
1
22
Q
i M QR
w
= Þ =w
p
18. (a) t = NBiA sinq so the graph between t and q is a
sinusoidal graph.
19. (d) Initial magnetic moment = m1 = iL2
I
m
12
 = L i
m
1
L
i
L/2
L
L/2
L
L M
M
m
2 M 2 =
After folding the loop, M = magnetic moment due to
each part = i
L
L
2
i
æö
´
ç÷
èø
 = 
2
L
2
i
 = 
1
µ
2
Page 2


1. ( b)
222
22
20 (4 10 ) 3 0.3
7
M NiA Am
-
= = ´ ´ ´=-
2. (a) The magnetic moment of current carrying loop
M = niA = ni (pr
2
)
Hence the work done in rotating it through 180°
W = MB(1 – cosq) = 2MB = 2(nipr
2
)B
= 2 × (50 × 2 × 3.14 ×16 × 10
–4
) × 0.1 = 0.1J
3. ( b)
(Number of turns)
NiAB
N
C
q = Þ qµ
4. ( d)
max
sin MB NiAB t= qÞt= , (q = 90°)
5. (c) In equilibrium angle between M
uur
 and B
ur
is zero. It
happens, when plane of the coil is perpendicular to B
ur
B
M
6. (a) t = NBiA = 100 × 0.2 × 2 × (0.08 × 0.1) = 0.32 N × m
7. (c) t = NBiA = 100 × 0.5 × 1 × 400 × 10
–4
 = 2 N–m
8. (a) t = NiAB sinq = 0 ( 0) q=° Q
9. (c) M = NiA Þ M µ A Þ M µ r
2
(As l = 2p r Þ lµ r )
Þ M µ l
2
10. (a)
11. (c)
2
max
1 () NiAB i rB t = =´´p´
2,
2
L
r Lr
æö
p = Þ=
ç÷
èø p
2 2
max
24
L L iB
iB
æö
t=p=
ç÷
èøpp
12. (b)
13. (b) The magnetic field at the centre of the solenoid is
B = m
0
 ni = (4p × 10
-7
) × (500/0.4) × 3 N/A.m.
The torque acting on a current-carrying coil having N
turns (say), placed perpendicular to the axis at the centre
of the solenoid is-
t = BiNA = (4p × 10
-7
) × (500/0.4) × 3 × 0.4 × 10
× p (0.01)
2
 = 6p
2
 × 10
-7 
= 5.92 × 10
-6
 N.m.
14. (d) The equivalent magnetic moment is
M = iA = ef (pr
2
)
As  f = 
v
2r p
\ M = 
ev
2r p
pr
2
 = 
evr
2
15. (b)
C
ii
NAB
q
= Þ µq
16. (d) Initially for circular coil L = 2pr and M = i × pr
2
= 
2 2
24
L iL
i
æö
´p=
ç÷
èøpp
     ...(i)
Finally for square coil
2 2
'
4 16
L iL
Mi
æö
=´=
ç÷
èø
    ...(ii)
L/4
i
Solving equation (i) and (ii) 
4
M
M
p
¢=
17. (b)
2
M iA iR = = ´p
also 
2
1
22
Q
i M QR
w
= Þ =w
p
18. (a) t = NBiA sinq so the graph between t and q is a
sinusoidal graph.
19. (d) Initial magnetic moment = m1 = iL2
I
m
12
 = L i
m
1
L
i
L/2
L
L/2
L
L M
M
m
2 M 2 =
After folding the loop, M = magnetic moment due to
each part = i
L
L
2
i
æö
´
ç÷
èø
 = 
2
L
2
i
 = 
1
µ
2
DPP/ P 41
116
Þ m
2
 = 
M2
 = 
1
µ
2
2
´ = 
1
µ
2
20. (a)
21. (d)
22. (a) Couple of force on loop S will be maximum because
for same perimeter the area of loop will be maximum
and magnetic moment of loop  = i ×A. So, it will also
be maximum for loop S.
23. (b) Sensitivity 
nBA
S
iC
q
==
24. (b)
mBsin t=q
 is zero for q = 0°, 180°.
25-27
25. (a) 26. (b) 27. (b)
(a) The net force on a current carrying loop of any
arbitrary shape in a uniform magnetic field is zero.
net
F0 =
r
(b) The given loop can be considered to be a
superposition of three loops as shown in figure. The
area vector of the three loops (1), (2) and (3) are
42
1
1
ˆ
A 10 10 10 jm
2
-
æö
= ´ ´´
ç÷
èø
r
42
2
1
ˆ
A 10 10 10 k m
2
-
æö
= ´´´
ç÷
èø
r
42
3
1
ˆ
A 10 10 10 i m
2
-
æö
= ´ ´´
ç÷
èø
r
Magnetic moment vector,
2
ˆ ˆˆ
iA 10(0.01i 0.005j 0.005k)Am m== ++
r
r
= 
2
ˆ ˆˆ
(0.1i 0.05j 0.05k)Am ++
c. Torque,
ˆ ˆ ˆ ˆ ˆˆ
(0.1i 0.05j 0.05k) (2i 3j k) t= + + ´ -+
r
= 
ˆ ˆˆ
i jk
ˆˆ
0.1 0.05 0.05 0.1i 0.4k Nm
2 31
= =--
-
28. (a) Due to metallic frame the deflection is only due to
current in a coil and magnetic field, not due to vibration
in the strings. If string start oscillating, presence of
metallic frame in the field make these oscillations
damped.
29. (b) The torque on the coil in a magnetic field is given by t
= nIBA cos q
For radial field, the coil is set with its plane parallel to
the direction of the magnetic field B, then q = 0° and
cos q = 1 Þ Torque = nIBA (1) = nIBA (maximum).
30. (c) Loop will not oscillate if in unstable equilibrium
position.