Question
Obtain an expression for the induced e.m.f. in a coil rotating with uniform angular velocity in uniform magnetic field. Plot a graph of variation of induceed e.m.f. against phase (0= wt) over one cycle.
✓
Answer
a) Consider a rectangular loop of conducting wire ‘PQRS’ partly placed in uniform
magnetic field of induction ‘B’ as shown in figure.
b) Let ‘l’ be the length of the side PS and ‘x’ be the length of the loop within the field.
∴ A = lx = area of the loop, which lies inside the field.
c) The magnetic flux (φ) through the area A at certain time ‘t’ is Φ = BA = Blx
d) The loop is pulled out of the magnetic field of induction ‘B’ to the right with a uniform
velocity ‘v’.
e) The rate of change of magnetic flux is given by, \(\frac{d \phi}{d t}=\frac{d}{d t}(B l x)\)
\(\therefore \quad \frac{ d \phi}{ dt }= B l\left(\frac{ dx }{ dt }\right)\)
But, \(\left(\frac{d x}{d t}\right)=v\)
\(\therefore \quad \frac{ d \phi}{ dt }= B / v\)...(i)
f) Due to change in magnetic flux, induced current is set up in the coil. The direction of
this current is clockwise according to Lenz’s law. Due to this, the sides of the coil
experience the forces, F1, F2 and F as shown in figure. The directions of these forces is
given by Fleming’s left hand rule.
g) The magnitude of force ‘F’ acting on the side PS is given by, F = BIl.
h) The force \(\vec{F}_1\) and \(\vec{F}_2\) are equal in magnitude and opposite in direction, therefore they cancel out. The only unbalanced force which opposes the motion of the coil is \(\vec{F}\) Hence, work must be done against this force in order to pull the coil.
i) The work done in time ‘dt’ during the small displacement ‘dx’ is given by, dW = −Fdx
− ve sign shows that F and ‘dx’ are opposite to each other.
∴ dW = − (BIl) dx ….(ii)
j) Mechanical power is given by,
\(P =\frac{ dW }{ dt }= BI l\left(\frac{ dx }{ dt }\right)\)
\(\therefore \quad P = BI / v \quad\left[\because \frac{ dx }{ dt }= v \right]\)
k) This external work provides the energy needed to maintain the induced current I
through the loop (coil).
l. If ‘e’ is the e.m.f induced then,
electric power \(=\frac{d W}{d t}=e I\)
∴ dW = eIdt .…(iii)
m) From equations (ii) and (iii), we have, eIdt = − BIl dx
\(\therefore e=-B l \frac{d x}{d t}\)
\(\therefore e =- B / v \ldots\) (iv)
n. From equation (i) and (iv), we have,
\(e=-\frac{d \phi}{d t}\)
o) Graph of variation of induced e.m.f. against phase (θ = ωt) over one cycle
magnetic field of induction ‘B’ as shown in figure.
b) Let ‘l’ be the length of the side PS and ‘x’ be the length of the loop within the field.
∴ A = lx = area of the loop, which lies inside the field.
c) The magnetic flux (φ) through the area A at certain time ‘t’ is Φ = BA = Blx
d) The loop is pulled out of the magnetic field of induction ‘B’ to the right with a uniform
velocity ‘v’.
e) The rate of change of magnetic flux is given by, \(\frac{d \phi}{d t}=\frac{d}{d t}(B l x)\)
\(\therefore \quad \frac{ d \phi}{ dt }= B l\left(\frac{ dx }{ dt }\right)\)
But, \(\left(\frac{d x}{d t}\right)=v\)
\(\therefore \quad \frac{ d \phi}{ dt }= B / v\)...(i)
f) Due to change in magnetic flux, induced current is set up in the coil. The direction of
this current is clockwise according to Lenz’s law. Due to this, the sides of the coil
experience the forces, F1, F2 and F as shown in figure. The directions of these forces is
given by Fleming’s left hand rule.
g) The magnitude of force ‘F’ acting on the side PS is given by, F = BIl.
h) The force \(\vec{F}_1\) and \(\vec{F}_2\) are equal in magnitude and opposite in direction, therefore they cancel out. The only unbalanced force which opposes the motion of the coil is \(\vec{F}\) Hence, work must be done against this force in order to pull the coil.
i) The work done in time ‘dt’ during the small displacement ‘dx’ is given by, dW = −Fdx
− ve sign shows that F and ‘dx’ are opposite to each other.
∴ dW = − (BIl) dx ….(ii)
j) Mechanical power is given by,
\(P =\frac{ dW }{ dt }= BI l\left(\frac{ dx }{ dt }\right)\)
\(\therefore \quad P = BI / v \quad\left[\because \frac{ dx }{ dt }= v \right]\)
k) This external work provides the energy needed to maintain the induced current I
through the loop (coil).
l. If ‘e’ is the e.m.f induced then,
electric power \(=\frac{d W}{d t}=e I\)
∴ dW = eIdt .…(iii)
m) From equations (ii) and (iii), we have, eIdt = − BIl dx
\(\therefore e=-B l \frac{d x}{d t}\)
\(\therefore e =- B / v \ldots\) (iv)
n. From equation (i) and (iv), we have,
\(e=-\frac{d \phi}{d t}\)
o) Graph of variation of induced e.m.f. against phase (θ = ωt) over one cycle
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