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Illustration:
Q1. A change q moves in a circular path of radius r with a speed of v. Calculate the induction of the magnetic field produced at the centre of the circle.
Solution: The equivalent current in the circular path is
I = 
B = 
Q2. The coil of a galvanometer has 500 turns and each turns has an avg. area 3 x 10-4 m2. calculate the magnetic moment of the coil when a current of 0.5 A passes throgh it. If a torque of 1.5 Nm is required for this coil carrying same current to set it parallel to a magnetic field. Calculate the magnetic field.
Solution: µ = NIA
= 500 x 0.5 x 3 x 10-4
= 0.075 Am2
Also  or | | = µB sin
where = angle between B and A
Here = 900
  = µB sin900
B =  = 20 T
ASSIGNMENTS:
Easy:
E1. Find the magnetude of magnetic field at point P due to the current carrying wire as shown.
Solution:
1
= - 30
0
,
2
= 60
0
E2. Shown infigure is a conductor carrying current I. Find the magnetic field intensity at the point O.
Solution: B =
[Due to an are]
E3. What is the work done in transferring the wire from position (1) to position(2) ?
Solution:
The 'd' magnetic moment of the loop = I
2
l dr
B
r
=
U
r
=
-
U = work done
- (U
2
- U
1
)
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Vriti Edustore
