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15. Your upright
thumb shows the
direction of the
magnetic force on
a positive particle.
Point your fingers
in the direction of
๐ and the curl
them toward the
direction of ๐ฉ.
44. Magnetic force on a Wire Carrying a current
F = I(L x B)
dF = I(dL x B)
F = โฎ ๐๐ = Iโฎ ๐ ๐ณ x B
CASE 1 : Straight wire
CASE 2 : Wire of any arbitrary wire
F = Iโฎ ๐๐ฅ x B
45. TORQUE ON A CURRENT
LOOP IN MAGNETIC
FIELD - 1
46.
47.
48.
49. TORQUE ON A CURRENT
LOOP IN MAGNETIC
FIELD - 2
50.
51.
52. TORQUE ON A CURRENT
LOOP IN MAGNETIC
FIELD - 3
78. Q. A coil of moving coil galvanometer has 100 turns and effective area 0.05๐๐ .
It is suspended in a radial magnetic field of induction 0.01 Wb/๐๐. The
torque per unit twist of the suspension fibre is 5 ร ๐๐โ๐ Nm/degree. What is
the current sensitivity of the moving coil galvanometer?
a) ๐๐๐
degree/ampere b)๐๐๐
degree/ampere
c) 2ร๐๐๐ degree/ampere d)3ร ๐๐๐ degree/ampere
94. MAGNETIC POTENTIAL ENERGY FOR A DIPOLE
Work done,
๐ = ๐ฆ๐ ๐ฌ๐ข๐ง๐
PE = โซ ๐
ds
๐ d๐ฝ
๐1
๐2
๐ ๐๐ณ
=
โซ
๐ฝ๐
๐ฝ๐
๐Bsin๐ฝ ๐ ๐ฝ mBโซ
๐ฝ๐
๐ฝ๐
๐๐๐๐ฝ ๐ ๐ฝ mB(โcos๐ฝ)๐ฝ๐
๐ฝ๐
= = =
U = -mb(cos๐ฝ๐.cos๐ฝ๐)
95. U = -mb(cos๐ฝ๐.cos๐ฝ๐)
U = -mb(cos๐ฝ๐.cos๐ฝ๐)
๐ฝ๐= ๐๐ยฐ ๐ฝ๐ = 0
U = -mB cos๐ฝ
U = -mB cos๐ฝ
Case 1 : ๐ฝ = ๐
Case 2 : ๐ฝ = ๐๐
Case 3 : ๐ฝ = ๐๐๐
U = -mB cos๐
U = -mB cos๐๐
U = -mB cos๐๐๐
U = -mB
U = 0
U = mB
MIN MAX
124. (4)
(5)
When point P lies far away from the center of coil (loop) then x >>> ๐,
Note
In each case, magnetic induction is directed along
the axis of the coil and perpendicular to the plane of
126. Magnetic Lines for a Current Loop:
We know that the magnetic field at a point P on the axis is
given by Eq. (10.53) as
๐ต๐ง =
๐0๐ผ๐ 2
2 z2 + ๐ 2 3/2
As a special case, the field at the centre of the loop is
obtained from the above equation by letting ๐ง = 0 :
๐ต0 =
๐0๐ผ
2๐
For a coil of N turns,
๐ต =
๐0๐๐ผ
2๐
The magnetic field lines from a circular loop are depicted
in Fig. 10.20. The direction of the field is as per the right
hand thumb rule: Curl the palm of your right hand along
the circular wire with the fingers in the direction of the
current. The stretched right hand thumb then gives the
direction of the magnetic field (Fig. 10.21). Thus, the
upper part of the loop seen in Fig. 10.20 may be regarded
as the North pole and the lower part as the South pole of a
bar magnet.
127. The magnetic moment ๐ of a circular loop is defined
as ๐ = ๐ผ๐ด, where ๐ด is a vector of magnitude ๐ด and
direction perpendicular to ๐ด. Using Eq. (10.56),
โด ๐ต๐ง =
๐0
2๐
๐
๐ง3
๐ต๐ง =
๐0
4๐
2๐
๐ง3
Note that ๐ต๐ง and ๐ are in the same direction,
perpendicular to the plane of the loop.
Using electrostatic analogue,
๐ธ =
2๐
4๐๐0๐ง3
,
which is the electric field at an axial point of an
electric dipole.
134. Q. Five very long insulated straight wire are bound together to form a small
cable. Current carried by the wires are I1 = 20A, I2 = -5A, I3 = 10A, I4 = 7A
and I5 = -12A. What is the magnetic induction at a perpendicular distance of 5
cm from the cable?
a) 60ยตT b) 70ยตT
c) 75ยตT d) 80ยตT
140. Inside the solenoid
the magic field is
uniform and magnetic
field lines are parallel
to the solenoid axis.
As ๐ = ๐๐
๐๐จ๐ฌ ๐๐ = 1