Example 6. In the middle of the long straight solenoid, which has 10 turns per centimeter, there is the circular contour with area 10 m 2. Plane of contour is situated at an angle of 300 to an axis of the solenoid (look at the picture). A current 5 A flows over the winding of solenoid. What average EMF is appears in the circular contour, if during 10 ms the direction of a current in solenoid is changed on the opposite one?
| Input data: n 0 = 10 cm 1 = 103 m 1 S = 10 cm 2 = 10 3 m 2; j = 30 0; =5; D t = 10 ms = 10 2 s. | 
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Find:  ?
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Solution: When a current in the coil of solenoid is changed, the magnetic flux of F, that runs through the circular contour, will change too. EMF of induction, which can be found in Faraday's law of electromagnetic induction e = d f / dt appears owing to this in contour. They find instantaneous value of EMF of induction with this formula, but for finding the average value from the average speed of magnetic flux changing in course of time we should write
(1)
where 
- the interlinkage changing in time interval of D t. Circular contour has a one loop N= 1. In the middle part of long solenoid field will be homogeneous. In this case the magnetic flux is found with the formula: = BS× cosa, where the solenoid magnetic field induction; S area of contour; α = 900j angle between vector 
and normal 
to the plane of contour.
The magnetic field induction is in the middle of long solenoid
(2)
where 
a magnetic constant; m magnetic permeability of medium (in air m=1) I current strength in the winding of solenoid; 
quantity of turns per the unit of length of a solenoid.
Lets choose the direction of a current in the winding of solenoid, as its shown in the picture I 1, and well find corresponding to the direction of vector 1 with the screw rule. Magnetic flux equals in this case
(3)
After changing the current direction on opposite one and vector of magnetic induction will change its direction 2 ¯ 1.
Magnetic Flux
. (4)
Here it is necessary to take into account, that a2=1800a and cosa2= cosa. Inserting the expressions F1 (3) and F2 (4) in the formula (1) and taking into account, that a=900j and cosa= cosj (look at the picture), we get
Lets check if the right part of the calculating formula gives the unity of EMF difference [ V ]:
Lets make the calculations
Result: 
Example 7 In the homogeneous magnetic field with induction of 0,5 short coil is uniformly rotated. Coil consists of 100 loops with the area of 100 m 2. The axis of rotation is in the plane of loops of the coil and perpendicularly to the lines of magnetic field induction (see the picture).
Find: 1) maximal EMF, which appears into coil with its rotation with period of 0.1 s;
2) charge, which flows over the coil within the change of angle between coils axis and magnetic inductions vector from j1=0o to j2=90o, if the coil resistance is 2W.
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Input Data:
=0,5 ,
N =100,
S =100 cm  ,
=0,1 s,
φ1=0, φ2=90
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| Find: emax; Q ? |
Solution: The magnetic flux which runs through the coil changes with the rotation of the coil, and in accordance with Faraday's law of electromagnetic induction in coil EMF of induction appears
(1)
where the magnetic flux interlinkage in the homogeneous magnetic field with induction of is found with the formula:
f= BSN× cosj (2)
where S area of loops; j angle between normal 
to the plane of coils loops and vector ; N quantity of loops, which are transpierced with the flux = BS× cosj. Coil during time of t will be turned an angle of j = w t, when the rotation is uniform, w angular velocity of rotation, which related with the period of rotation as 
Substituted expressions w and j in the formula (2) well get
f = BSN× cosw t = 
. (3)
In order to find the instantaneous value of EMF of induction well substitute the expression for magnetic flux interlinkage (3) in the formula (1) and differentiate with time
The value of EMF will be maximum at that moment of time, if 
Then 
. (4)
In order to find the charge of Q, which runs over the coil, lets apply Ohm's law for a closed circuit instantaneous value of EMF
eI= Ri (5)
where R coils resistance; i instantaneous value of current in the coil.
Equating the right parts of equations (1) and (5), well get
As there is instantaneous value of current strength i=dQ / dt then this equation can be rewritten in the type of
whence
(6)
If one takes integral from expression (6), then well find charge, which flows through the coil with the change of magnetic flux interlinkage from f1 till f2:
or 
(7)
Changing the angle between coils axis and vector from j1 to j2, then the magnetic flux interlinkage will change from f1= BSN× cosj1 till f2= BSN× cosj2. Substituted expressionsf 1 and f2 in the formula (7), well find the sought-for charge:
. (8)
Lets check if the right part of the formula gives (4) the unit of EMF (), and the right part of the formula (8) the unit of charge (C):
;
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Lets make the calculations:
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Result: 
, 
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Individual tasks for PROBLEM 2.4.
