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
)