The angle \(\theta\) made by the area vector of the coil with the magnetic field is 45° . From Eq. (6.1), the initial magnetic flux is
\(\phi\) = BA cos \(\theta\)
\(\begin{aligned} =\frac{0.1 \times 10^{-2}}{\sqrt{2}} \mathrm{Wb} \end{aligned}\)
Final flux, F_min = 0
The change in flux is brought about in 0.70 s. From Eq. (6.3), the magnitude of the induced emf is given by
\(\varepsilon=\frac{\left|\Delta \Phi_B\right|}{\Delta t}=\frac{|(\Phi-0)|}{\Delta t}=\frac{10^{-3}}{\sqrt{2} \times 0.7}=1.0 \mathrm{mV}\)
And the magnitude of the current is
\(I=\frac{e}{R}=\frac{10^{-3} \mathrm{~V}}{0.5 \Omega}=2 \mathrm{~mA}\)
Note that the earth’s magnetic field also produces a flux through the loop. But it is a steady field (which does not change within the time span of the experiment) and hence does not induce any emf.

A) Step 1: Find the emf developed in the loop
Formula Used: \(\mathrm{e}=\mathrm{BIV}\)
Strength of magnetic field, \(B=0.3 \mathrm{~T}\)
Velocity of the loop, \(v=1 \mathrm{~cm} / \mathrm{s}=0.01 \mathrm{~m} / \mathrm{s}\)
emf developed in the loop is given as:
\( \mathrm{e}=\mathrm{Blv} \)
\(=0.3 \times 0.08 \times 0.01=2.4 \times 10^{-4} \mathrm{~V}\)
Step 2: Find the time taken to travel.
Formula Used: \(\mathrm{t}=\frac{\text { Distance travelled }}{\text { Velocity }}\)
Time taken to travel along the width, \(\mathrm{t}=\frac{\text { Distance travelled }}{\text { Velocity }}=\frac{\mathrm{bv}}{\mathrm{v}}\)
\(=\frac{0.02}{0.01}=2 \mathrm{~s}\)
Final answer : \(\mathrm{e}=2.4 \times 10^{-4} \mathrm{v}\)
\(\mathrm{t}=2 \mathrm{~s}\)
B) Step 1: Find the emf developed in the loop emf developed, e = BIv
\(e=0.3 \times 0.02 \times 0.01 \)
\(e=0.6 \times 10^{-4} V\)
Step 2: Find the time taken to travel.
Time taken to travel along the length, \(\mathrm{t}=\frac{\text { Distance travelle }}{\text { Velocity }}=\frac{\mathrm{d}}{\mathrm{v}}\)
\(\mathrm{t}=\frac{0.08}{0.01}=8 \mathrm{~s}\)
Hence, the induced voltage is \(0.6 \times 10^{-4} \mathrm{~V}\) which lasts for 8 s .
Final answer: \(\mathrm{e}=0.6 \times 10^{-4} \mathrm{~V}\)
\(\mathrm{t}=8 \mathrm{~s}\)

Yes, current induced, \(I=\frac { N }{ R } \frac { d\phi }{ dt } \)

The back emf in the coil opposes the change in the current as per Lenz's law.

Induced emf  = (1/2)\(\omega\)BR²
= (1/2) \(\times\)4\(\pi\)\(\times\)0.4\(\times\)10^-4 \(\times\)(0.5)²
= 6.28 \(\times\)10^-5 V
The number of spokes is immaterial because the emf’s across the spokes are in parallel.

The direction of the induced current in a closed loop is given by Lenz’s law. The given pairs of figures show the direction of the induced current when the North pole of a bar magnet is moved towards and away from a closed loop respectively.

Using Lenz’s rule, the direction of the induced current in the given situations can be predicted as follows:
(a) The direction of the induced current is along qrpq.
(b) The direction of the induced current is along prqp.
(c) The direction of the induced current is along yzxy.
(d) The direction of the induced current is along zyxz.
(e) The direction of the induced current is along xryx.
(f) No current is induced since the field lines are lying in the plane of the closed loop.

Method I
As the rod is rotated, free electrons in the rod move towards the outer end due to Lorentz force and get distributed over the ring. Thus, the resulting separation of charges produces an emf across the ends of the rod. At a certain value of emf, there is no more flow of electrons and a steady state is reached. the magnitude of the emf generated across a length dr of the rod as it moves at right angles to the magnetic field is given by
dε = Bv dr . Hence,
\(\varepsilon = \int \mathrm{~d} \varepsilon=\int_0^R B v \mathrm{~d} r=\int_0^R B \omega r \mathrm{~d} r=\frac{B \omega R^2}{2}\)
Note that we have used v = ω r. This gives
\(\varepsilon=\frac{1}{2} \times 1.0 \times 2 \pi \times 50 \times\left(1^{2}\right)\)
= 157 V
Method II
To calculate the emf, we can imagine a closed loop OP\(\mathcal{Q}\) in which point O and P are connected with a resistor R and O\(\mathcal{Q}\) is the rotating rod. The potential difference across the resistor is then equal to the induced emf and equals B x (rate of change of area of loop). If θ is the angle between the rod and the radius of the circle at P at time t, the area of the sector OP\(\mathcal{Q}\) is given by
\(\pi R^{2} \times \frac{\theta}{2 \pi}=\frac{1}{2} R^{2} \theta\)
where R is the radius of the circle. Hence, the induced emf is
\(\varepsilon=B \times \frac{\mathrm{d}}{\mathrm{d} t}\left[\frac{1}{2} R^{2} \theta\right]=\frac{1}{2} B R^{2} \frac{\mathrm{d} \theta}{\mathrm{d} t}=\frac{B \omega R^{2}}{2}\)
\(\text { [Note: } \left.\frac{\mathrm{d} \theta}{\mathrm{d} t}=\omega=2 \pi v\right]\)
This expression is identical to the expression obtained by Method I and we get the same value of ε.

(a) No. However strong the magnet may be, current can be induced only by changing the magnetic flux through the loop.
(b) No current is induced in either case. Current can not be induced by changing the electric flux.
(c) The induced emf is expected to be constant only in the case of the rectangular loop. In the case of circular loop, the rate of change of area of the loop during its passage out of the field region is not constant, hence induced emf will vary accordingly.
(d) The polarity of plate ‘A’ will be positive with respect to plate ‘B’ in the capacitor.

Initial flux through the coil,
Φ_{B (initial)} = BA cos θ
= 3.0 x 10^–5 x (π x 10^–2) x cos 0º
= 3π x 10^–7 Wb
Final flux after the rotation,
Φ_{B (final)} = 3.0 x 10^–5 x (π x 10^–2) x cos 180°
= –3π x 10^–7 Wb
Therefore, estimated value of the induced emf is,
\(\varepsilon=N \frac{\Delta \Phi}{\Delta t}\)
= 500 x (6π x 10^–7)/0.25
= 3.8 x 10^–3 V
I = ε/R = 1.9 x 10^–3 A
Note that the magnitudes of ε and I are the estimated values. Their instantaneous values are different and depend upon the speed of rotation at the particular instant.

Lenz's law. It states that the direction of induced current (or e.m.f.) is such that it always opposes the cause producing it. This law gives the direction of an induced current in terms of the cause of the current. An induced current in a coil is always due to change of magnetic flux linking the coil. The induced current will have its own associated magnetic flux of induced current will be such as to oppose the change of the primary magnetic flux.
Experimental verification of Lenz's law:
Let us connect the two ends of a coil of few turns to a cell C and a galvanometer G through a two-way key having terminals 1, 2 and 3 as shown in Fig. Put the plug between gap 1 and 2, the current in the upper face of the coil is in anticlockwise direction of N-pole is produced on this upper face. Let the galvanometer shows deflection to the right side. Now remove the key from gaps 1 and 2 and insert between 2 and 3, so the cell C is cut off. Bring N-pole of a bar magnet towards the upper face of the coil, the deflection produced in the galvanometer is again towards right. It means that the upper face of the coil becomes N-pole and repels the incoming magnet. Now take the magnet away from the coil, the deflection in the galvanometer is now towards left i.e. S-pole is formed at the upper end of the coil and it attracts the N-pole of the magnet. 

Thus we find that whether the magnet is taken towards or away from the coil, the direction of induced e.m.f. is such that it always opposes the motion of the magnet as predicted by Lenz's law.

(i) The magnetic flux through the rectangular loop abcd increases, due to the motion of the loop into the region of magnetic field, The induced current must flow along the path bcdab so that it opposes the increasing flux.
(ii) Due to the outward motion, magnetic flux through the triangular loop abc decreases due to which the induced current flows along bacb, so as to oppose the change in flux.
(iii) As the magnetic flux decreases due to motion of the irregular shaped loop abcd out of the region of magnetic field, the induced current flows along cdabc, so as to oppose change in flux. Note that there are no induced current as long as the loops are completely inside or outside the region of the magnetic field

No because, \(\Phi=N B A=\text { constant }\)
\(\therefore e=\frac{d \Phi}{d t}=0 ; i=0\)

No, Lenz's law does not violate this principle.

Change in magnetic flux linked with the circuit.

This is because induced e.m.f. opposes the current due to the actual source e.m.f.

When plane of coil is along the field.

Weber, Tesla.

Yes, it will appear as the train is intercepting vertical component of earth's magnetic field. No, answer is not affected.

In the first case, induced current in B will be opposite to i, so that A and B repel eachother and the operation of bringing them closer is opposed. Similarly, in the second case, induced current in B will be in the direction of i.

No, because the pole intercepts neither H nor V.

Due to change in magnetic flux, e.m.f. is always induced, but induced current will flow only when tghe circuit is complete.

\(|e|=\frac{d \Phi}{d t}=\frac{0.2 \times 10^{-2}}{0.1}=0.02 \quad V\)

1 weber = 10⁸ Maxwell

\(\Phi=B A \cos \Phi=\left(\frac{F}{I l}\right) A=\frac{M L T^{-2}}{A L} \cdot L^2\)
\(=\left[M L^2 T^{-2} A^{-1}\right]\)

The magnetic flux is maximum when area is held perpendicular to the direction of magnetic field.