A Quantitative Treatment
To put Faraday’s law to work. we need a way to calculate the amount of magnetic field that passes through a loop. In Chapter 24, in a similar situation, we needed to calculate the amount of an electric field that passes through a surface. There we defined an electric flux ¢lE = f i·iA. Here we define a magnetic flux: Suppose aloop enclosing an area A is placed in a magnetic field ii. Then the magnetic flux through the loop is
<JIB = f t.iA (magnetic flux through area A).
As in Chapter 24, dA is a vector of magnitude dA that is perpendicular to a differential area dA.
As a special case of Eq. 31-3, suppose that the loop lies in a plane and that the magnetic field is perpendicular to the plane of the loop. Then we can write the dot product in Eq. 31-3 as B dA cos 0° = B dA. If the magnetic field is also uniform, then B can he throught out in front of the integral sign. The remaining f dA then
Sample Problem 31-1
The long solenoid S shown (in cross section) in Fig. 31-3 has 220 turns/em and carries a current i = J.5 A; its diameter D is 3.2 cm.At its center we place a J30-turn closely packed coil C of diameterd = 2. J cm. The current in the solenoid is reduced to zero at a teady rate in 25 ms. What is the magnitude of the emf that is induced in coil C while the current in the solenoid is changing? SOLUTION: The Key Ideas here are these: noid S. which carries current i.
I. Because coil C is located in the interior of the solenoid. it lies
within the magnetic field produced by current i in the solenoid;
thus. there is a magnetic flux <1>8 through coil C.
2. Because current i decreases. nux <1>8 also decreases.
3. As <1>8 decreases, emf ~ is induced in coil C. according to Faraday’s
Because coil C consists of more than one turn. we apply Faraday’s law in the form of Eq. 31-7 (& = – N d<l>8/dr). where the number of turns N is 130 and d<l>8/dr is the rate at which the flux in each
turn changes. Because the current in the solenoid decreases at a steady rate.flux <1>8 also decreases at a steady rate and we can write d<J>8/dr a~ .l<1>8/<lr. Then. to evaluate .l<l>8′ we need the final and initial flux.The final flux <1>8./ is zero because the final current in the solenoid is zero. To find the initial flux <J>8.•• we need two more Key Ideas: . The flux through each turn of coil C depends on the area A and
orienta ion of that turn in the. solenoid’s magnetic field if. Because if is uniform and directed perpendicular to area A. the flux is given by Eq. 31-4 (<1>8 = BA). 5. The magnitude B of the magnetic field in the interior of a solenoid
depends on the solenoid’s current i and it number n of turns per unit length. according to Eq. 30-25 (B = /Loin).’
For the situation of Fig. 31-3. A i~ 117d2 (= 3.46 x 10-4 m’) and n is 220 turns/em, or 22 000 turns/m. Substituting Eq. 30-25 into Eq. 31-4 then leads to “‘0., = BA = (/Loin)A = (417 X 10 7 T· m/A)(1.5 A)(22 ()()( turns/m)
X (3.46 X 10-4 m2) = 1.44 X IO-~ Wb. Now we can write
d<ll8 = .l<J>8 = <J>8.( – <1>/1., dr.lr l.r
(0 – 1.44 X IO-~ Wb)
25 X 10-” s
= -5.76 X 10-4 Wb/s = -5.76 X 10 4 V.
We are interested only in magnitudes. so we ignore the minus signs
here and in Eq. 31-7. writing
‘& = N _8 = (130 turns)(5.76 X 10-4 V)
= 7.5 X 10-1 V = 75 mY. (Answer)