Question
Obtain an expression for e.m.f. induced in a coil rotating with uniform angular velocity in a uniform magnetic field. Show graphically the variation of e.m.f. with time (t).
✓
Answer
Coil rotating in uniform magnetic field:
Consider a coil of ‘N’ turns with effective area NA placed in uniform magnetic induction B as shown in the figure below.
The coil is rotated continuously with constant angular velocity ω. The axis of rotation is in the plane of the coil and normal to the magnetic induction B.
At t = 0, the plane of the coil is perpendicular to the magnetic induction B.
The magnetic flux passing through the coil is NAB.
After t seconds, the plane of the coil is at an angle θ.
Thus, the magnetic flux Φ through the coil at time t is given by
Φ = NABcosθ = NABcosωt
As time changes, the magnetic flux goes on changing. Hence, the e.m.f. generated in the coil is given by
\(e=-\frac{d \phi}{d t}=-\frac{d}{d t}(N A B \cos \omega t)\)
e = NABωsinωt
e = 2πfNABsinωt
This is the expression for induced e.m.f. generated in the coil at any instant t. It is known as instantaneous e.m.f.
Given that
$R/L = \sigma = 0.1Ω/cm = 0.1 x 100 = 10Ω/m$
$l_1 = 300 cm = 3 m, E_1 = 1.5 V, E_2= 1.4 V$
We know that
$E_1 = iR_1 =i\sigma l_1$
\(\therefore i=\frac{E_1}{\sigma l_1}=\frac{1.5}{10 \times 3}=0.05 A\)
Hence, the current for the other cell is 0.05 A.
For potentiometer, the balancing condition is
\(\frac{E_1}{E_2}=\frac{l_1}{l_2}\)
\(\therefore l_2=l_1 \frac{E_2}{E_1}=3 \times \frac{1.4}{1.5}=2.8 m\)
So, the balancing length for the other cell is 2.8 m.
Consider a coil of ‘N’ turns with effective area NA placed in uniform magnetic induction B as shown in the figure below.
The coil is rotated continuously with constant angular velocity ω. The axis of rotation is in the plane of the coil and normal to the magnetic induction B.
At t = 0, the plane of the coil is perpendicular to the magnetic induction B.
The magnetic flux passing through the coil is NAB.
After t seconds, the plane of the coil is at an angle θ.
Thus, the magnetic flux Φ through the coil at time t is given by
Φ = NABcosθ = NABcosωt
As time changes, the magnetic flux goes on changing. Hence, the e.m.f. generated in the coil is given by
\(e=-\frac{d \phi}{d t}=-\frac{d}{d t}(N A B \cos \omega t)\)
e = NABωsinωt
e = 2πfNABsinωt
This is the expression for induced e.m.f. generated in the coil at any instant t. It is known as instantaneous e.m.f.
Given that
$R/L = \sigma = 0.1Ω/cm = 0.1 x 100 = 10Ω/m$
$l_1 = 300 cm = 3 m, E_1 = 1.5 V, E_2= 1.4 V$
We know that
$E_1 = iR_1 =i\sigma l_1$
\(\therefore i=\frac{E_1}{\sigma l_1}=\frac{1.5}{10 \times 3}=0.05 A\)
Hence, the current for the other cell is 0.05 A.
For potentiometer, the balancing condition is
\(\frac{E_1}{E_2}=\frac{l_1}{l_2}\)
\(\therefore l_2=l_1 \frac{E_2}{E_1}=3 \times \frac{1.4}{1.5}=2.8 m\)
So, the balancing length for the other cell is 2.8 m.
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