Electromagnetic Induction - Class 12 Engineering Physics - Extra Questions

SI unit of inductance is Henry denoted by $$H$$.

Faraday discovered that when he moved a magnet near a wire a voltage was generated across it. If the magnet was held stationary no voltage was generated, the voltage only existed while the magnet was moving. We call this voltage the induced emf $$E$$.

Faraday's Law of electromagnetic induction

The emf $$E$$ produced around a loop of the conductor is proportional to the rate of change of the magnetic flux (φ) through the area (A) of the loop.
$$E= -\dfrac {\mathrm{d \phi}}{dt}$$

Faraday's Electromagnetic Induction state that whenever a conductor are placed in a varying magnetic field emf are induced which is called induced emf, if the conductor circuit are closed current are also induced which is called induced current. The induced emf is equal to rate of change of flux linkages. 

The henry $$($$symbolized $$H)$$ is the standard international $$(SI)$$ unit of inductance$$.$$ Reduced to base $$(SI)$$ units$$,$$ one henry is the equivalent of one kilogram meter squared per second squared per ampere squared $$\left( {kg\,{m^2}{s^{ - 2}}{A^{ - 2}}} \right).$$

Hint: This law describes how a magnetic field interacts with an electric circuit to produce an electromotive force (emf).
 Solution: Step 1: Concept used: Faraday's first law of electromagnetic induction states that whenever the flux of magnetic field through the area bounded by closed-loop changes, an emf is produced in the loop.
Step 2: We can change the magnetic field in several ways: (a) moving a bar magnet towards and away from the coil. (b)The coil's movement in and out of a magnetic field. (c) Changing the area in which it positioned the coil, resulting in a change in the magnetic flux. (d) The magnetic field will vary according to the coil's rotation relative to the magnet.

Let us consider a coil of wire connected to a galvanometer, which we can use to measure current in the coil. There is no battery or power supply, so no current should flow. Now a magnet is brought close to the coil.
We notice two things:

• If the magnet is held stationary near, or even inside, the coil, no current will flow through the coil.
• If the magnet is moved, the galvanometer needle will deflect, showing that current is flowing through the coil.

When the magnet is moved one way (say, into the coil), the needle deflects one way; when the magnet is moved the other way (say, out of the coil), the needle deflects the other way. The direction of the current depends on how the magnet is moved. It seems like a constant magnetic field does nothing to the coil, while a changing field causes a current to flow.
To confirm this, the magnet can be replaced with a second coil, and a current can be set up in this coil by connecting it to a battery. The second coil acts just like a bar magnet. When this coil is placed next to the first one, which is still connected to the galvanometer, nothing happens when a steady current passes through the second coil. When the current in the second coil is switched on or off, or changed in any way, however, the galvanometer responds, indicating that a current is flowing in the first coil.
We notice one more thing. If we squeeze the first coil, changing its area, while it's sitting near a stationary magnet, the galvanometer needle moves, indicating that current is flowing through the coil.
We can conclude from all these observations that a changing magnetic field will produce a voltage in a coil, causing a current to flow. To be completely accurate, if the magnetic flux through a coil is changed, a voltage will be produced. This voltage is known as the induced emf.

(a) It is safer to be inside a car during thunderstorm because the car acts like a Faraday cage.
(b)Awareness and humanity
(c) Gratitude and obliged
(d) Once I came across to a situation where a puppy was struck in the middle of a busy road during rain and was not able to cross due to heavy flow, so I quickly rushed and helped him.

The angle the plane of coil makes with the magnetic field is $$\theta=\omega t$$.

Consider a coil of cross section area $$S$$ rotating with an angular speed $$\omega$$ in a uniform magnetic field $$B$$ as shown in the figure.

Principle : Single-phase a.c generator is based on the principle of electromagnetic induction, according to which an emf is induced in a coil when it is rotated in a uniform magnetic field.
Essential parts of an AC generator:
(i) Armature : It is a rectangular coil consisting of a large number of loops or turns of insulated copper wire wound over a laminated soft iron core or ring. The soft iron core not only increases the magnetic flux but also serves as a support for the coil.
(ii) Field magnets : The necessary magnetic field is provided by permanent magnets in the case of low power dynamos. For high power dynamos, field is provided by electromagnet. Armature rotates between the magnetic poles such that the axis of rotation is perpendicular to the magnetic field.
(iii) Slip rings : The ends of the armature coil are connected to two hollow metallic rings $$R_{1}$$ and $$R_{2}$$ called slip rings. These rings are fixed to a shaft, to which the armature is also fixed. When the shaft rotates, the slip rings along with the armature also rotate.
(iv) Brushes : $$B_{1}$$ and $$B_{2}$$ are two flexible metallic plates or carbon brushes. They provide contact with the slip rings by keeping themselves pressed against the ring. They are used to pass on the current from the armature to the external power line through the slip rings.
Working : Whenever, there is a change in orientation of the coil, the magnetic flux linked with the coil changes, producing an induced emf in the coil. The direction of the induced current is given by. Fleming's right-hand rule.
Suppose the armature $$ABCD$$ is initially in the vertical position. It is rotated in the anticlockwise direction. The side $$AB$$ of the coil moves downwards and the side $$DC$$ moves upwards (figure). Then according to Fleming's right-hand rule the current induced in arm $$AB$$ flows from $$B$$ to $$A$$ and in $$CD$$ it flows from $$D$$ to $$C$$. Thus the current flows along $$DCBA$$ in the coil. In the external circuit the current flows from $$B_{1}$$ to $$B_{2}$$.
On further rotation, the arm $$AB$$ of the coil moves upwards and $$DC$$ moves downwards. Now the current in the coil flows along $$ABCD$$. In the external circuit the current flows from $$B_{2}$$ of $$B_{1}$$. As the rotation of the coil continuous, the induced current in the external circuit keeps changing its direction for every half a rotation of the coil. Hence the induced current is alternating in nature (fig). As the armature completes $$n$$ rotations in one second, alternating current of frequency $$v$$ cycles per second is produced. The induced emf at any instant is given by $$e = E_{o}\sin \omega t$$.
The peak value of the emf, $$E_{o} = NBA\omega$$
where $$N$$ is the number of turns of the coil,
$$A$$ is the area enclosed by the coil,
$$B$$ is the magnetic field and
$$\omega$$ is the angular velocity of the coil.

Let  us  consider  a  straight  conductor  moving  in  a  uniform  and  time-independent magnetic field. 
The figure  shows a rectangular wire PQRS in which the conductor PQ is free to move. 
Rod PQ is moved towards the left with a constant velocity v.
PQRS forms closed circuit enclosing an area that changes as PQ moves.
It is placed in a uniform magnetic field B perpendicular to the plane of our system.
Let the length RQ =x and RS =l,
the magnetic flux $$\phi_B$$ enclosed  by the loop PQRS
$$\phi_B=Blx$$
As x is changing with time, the rate of change of flux $$\phi_B$$ will induce an emf given according to Faraday's law of electromagnetic induction  :
$$\epsilon=-\frac{d\phi_B}{dt}=-\frac{d}{dt} (Blx)=-Blv$$
This induced emf Blv is motional emf.

Here magnetic field is perpendicular to velocity the emf induce is (e).

There are two Faraday's laws of Electromagnetic induction,

Number of turns  $$N = 10000$$ turns
Area of coil  $$A = 100 \ cm^2 = 0.01 \ m^2$$
Frequency  $$f = 140 \ rpm = \dfrac{140}{60} \ rps$$
Angular speed  $$w = 2\pi f = 2\times \dfrac{22}{7}\times \dfrac{140}{60} = 14.7 \ rad/s$$
Magnetic field  $$B = 3.6\times 10^{-2} \ T$$
Maximum value of emf induced  $$\mathcal {E}_{max} = NBA w$$
$$\implies$$  $$\mathcal {E}_{max} = 10000\times 3.6\times 10^{-2}\times 0.01\times 14.7 = 52.9 \ $$ volts

Angular speed $$=50 \dfrac{rev}{sec}$$

According to Faraday's law of electromagnetic induction the magnitude of  induced EMF is equal to the rate of change of magnetic flux linked with the closed circuit or coil. Mathematically

$$v=1800k/h$$

Initial current passing $$=i$$
Hence initial emf $$=ir$$
Emf due to motion of $$ab=Blv$$
Net emf$$=ir-Blv$$
Net resistance $$=2r$$
Hence current passing $$=\cfrac { ir-Blv }{ 2r } $$

Force on the wire $$=ilB$$
Acceleration $$\cfrac { ilB }{ m } $$
Velocity $$=\cfrac { ilBt }{ m } $$

$$l=0.3m,\vec { B } =2.0\times { 10 }^{ -5 }T,\omega =100rpm$$
$$v=\cfrac { 100 }{ 60 } \times 2\pi =\cfrac { 10 }{ \pi  } rad/s$$
$$\quad v=\cfrac { l }{ 2 } \times \omega =\cfrac { 0.3 }{ 2 } \times \cfrac { 10 }{ 3 } \pi $$
$$emf=e=Blv=2.0\times { 10 }^{ -5 }\times 0.3\times \cfrac { 0.33 }{ 2 } \times \cfrac { 10 }{ 3 } \pi \quad =3\pi \times { 10 }^{ -6 }\times 0.3\times 3.14\times { 10 }^{ -6 }V=9.42\times { 10 }^{ -6 }V\quad \quad $$

(a) Fleming's right hand rule: Hold the thumb, the fore finger and the centre finger of your right-hand at right angles to one another. Adjust you hand in such a way that forefinger points in the direction of magnetic field, and thumb points in the direction of motion of conductor, then the direction in which centre finger points, gives the direction of induced current in the conductor.

(a) The circuit consists of one generator across one inductor; therefore,  $$\varepsilon _{m}=V_{L}$$. The current amplitude is
$$I_m=\dfrac{\varepsilon _{m}}{X_{L}}=\dfrac{\varepsilon _{m}}{\omega _{d}L}=\dfrac{25.0V}{\left ( 377rad/s \right )\left ( 12.7H \right )}=5.22 \times 10^{-3}A$$.

(b)$$I = I_m sin(\omega _d t - \phi)$$ --(i)

$$ϕ =LI $$ 

Change of flux for small change in current 
$$d\varphi =MdI =1.5(20-0)\ weber =30\ weber$$

Lenz's Law. The polarity of induced emf is such that it tends to produced a current which opposes the change in magnetic flux that produces it.

The direction of induced current in the wire loop when the magnet is moved towards it is in Anticlockwise from the side of a magnet.

The current induced in a closed circuit only if there is change in number of magnetic field lines linked with the circuit.

Here, $$e=200V$$;
$$dl=0.5=5A$$;
$$dt=0.1s$$
Now, $$e=-L\frac{dl}{dt}$$
Or, $$L=-\frac{e}{dl/dt}=-\frac{200}{-5/0.1}=4H$$

Here, $$M=1.5H$$;
$$dI=20-0=20A$$;
$$dt=0.5s$$;
Now, $$\varphi =MI$$
Or, $$d\varphi =MdI=1.5*20=30Wb$$

From the relation,
$$\phi =LI$$
It follows that the self inductance of a conductor is equal to the slope of its graph between $$\phi$$ and $$I$$. Therefore, the self inductance of the conductor $$A$$ has the larger value.

The flux linked to the ring can not change on transition to the supercondition state, for reasons, similar to that given above. Thus a current $$I$$ must be induced in the ring, where,
$$I = \dfrac {\Phi}{L} = \dfrac {\pi a^{2} B}{\mu_{0} a\left (ln \dfrac {8a}{b} - 2\right )} = \dfrac {\pi aB}{\mu_{0} \left (ln \dfrac {8a}{b} - 2\right )}$$.

(i) : Mutual inductance $$(M)$$ is the property of the coils that a current is induced in the secondary coil if there is a change in current in the primary coil.
Its SI unit is Henry.
(ii) :
Given :  $$M = 1.5 H$$    
Change of current  $${\Delta I}= {20-0} = 20 \ A$$
Change of flux  $$\Delta \phi =  M \Delta I = 1.5\times 20  =30 \ Wb$$

The diagram of AC generator is shown in the attached diagram.

(a)The emf is highest between diameter perpendicular to the velocity. Because here length perpendicular to velocity is highest
$${E}_{max}=VB2R$$
(b) The length perpendicular to velocity is lowest as the diameter is parallel to the velocity $${E}_{min}=0$$(a)The emf is highest between diameter perpendicular to the velocity. Because here length perpendicular to velocity is highest
$${E}_{max}=VB2R$$
(b) The length perpendicular to velocity is lowest as the diameter is parallel to the velocity $${E}_{min}=0$$

Hint : Use the concept of motional emf induced in a moving conductor rod to find emf induced in a rotating rod.

From $$t = 0 $$ to $$t = 3s (= \cfrac{30 \space cm}{10 \space cm/sec})$$ the flux through coil is zero.

$$=A\dfrac{\pi R^2}{2} +(2R\times x)$$
$$\phi =BA =\dfrac{B\pi P^2}{2}+(2R\times x)$$
$$\dfrac{d}{dt}$$
$$E=2BR(V)$$
$$E=(B)(v)(2R)$$
$$=2BvR$$

$$\quad I=\cfrac { Blv }{ R } =\cfrac { B\times l\cos { \theta  } \times v\cos { \theta  }  }{ R } =\cfrac { Blv }{ R } \cos ^{ 2 }{ \theta  } $$
$$=ilB=\cfrac { Blv\cos ^{ 2 }{ \theta  } \quad \times lB }{ R } $$
Now $$F=mg\sin { \theta  } $$ (Force due to gravity which pulls downwards)
Now $$\cfrac { { B }^{ 2 }{ l }^{ 2 }v\cos ^{ 2 }{ \theta  }  }{ R } =mg\sin { \theta  } \Rightarrow B=\sqrt { \cfrac { Rmg\sin { \theta  }  }{ { l }^{ 2 }v\cos ^{ 2 }{ \theta  }  }  } $$

$$l=20cm=0.2m;v=10cm/s=0.1m/s;B=0.10T$$
(a) $$F=qvB=1.6\times { 10 }^{ -19 }\times 1\times { 10 }^{ -1 }\times 1\times { 10 }^{ -1 }=1.6\times { 10 }^{ -21 }N$$
(b) $$qE=qvB\Rightarrow E=1\times { 10 }^{ -1 }\times 1\times { 10 }^{ -1 }=1\times { 10 }^{ -2 }V/m\quad $$
this is created due to the induced emf
(c) Motional $$emf=Bvl$$
$$=0.1\times 0.1\times 0.2=21\times { 10 }^{ -3 }V\quad $$

$$emf=\cfrac { 1 }{ 2 } B{ a }^{ 2 }\omega $$
current $$\cfrac { e+E }{ R } =\cfrac { 1/\times B{ a }^{ 2 }\omega +E }{ R } =\cfrac { B{ a }^{ 2 }\omega +2E }{ 2R } \quad \Rightarrow mg\cos { \theta  } =ilB$$ (net force actin on the rod is $$O$$)
$$\Rightarrow mg\cos { \theta  } =\cfrac { B{ a }^{ 2 }\omega +2E }{ 2R } a\times B\Rightarrow R=\cfrac { \left( B{ a }^{ 2 }\omega +2E \right) aB }{ 2mg\cos { \theta  }  } \quad \quad $$

The 2 resistances $$r/4$$ and $$3r/4$$ are in parallel
$$R'=\cfrac { r/4\times 3r/4 }{ r } =\cfrac { 3r }{ 16 } \quad \quad $$
$$e\quad e=Bvl=B\times \cfrac { a }{ 2 } \omega \times a=\cfrac { { B }^{  }{ a }^{ 2 }\omega  }{ 2 } $$
$$\quad i=\cfrac { e }{ R' } =\cfrac { { B }^{  }{ a }^{ 2 }\omega  }{ 2R' } =\cfrac { { B }^{  }{ a }^{ 2 }\omega  }{ 2\cfrac { 3r }{ 16 }  } =\cfrac { 8 }{ 3 } \cfrac { { B }^{  }{ a }^{ 2 }\omega  }{ r } $$

emf developed $$=Bdv$$ (When it attains speed $$v$$)
Current $$=\cfrac { Bdv }{ R } $$
Force $$=\cfrac { B{ d }^{ 2 }{ v }^{ 2 } }{ R } $$
This force opposes the given force
Net $$F=F-\cfrac { B{ d }^{ 2 }{ v }^{ 2 } }{ R } =RF-\cfrac { B{ d }^{ 2 }{ v }^{ 2 } }{ R } $$
Net acceleration $$=\cfrac { RF-{ B }^{ 2 }{ d }^{ 2 }{ v }^{ 2 } }{ R } $$
(b) Velocity becomes constant when acceleration is $$0$$
$$\cfrac { F }{ m } -\cfrac { { B }^{ 2 }{ d }^{ 2 }{ v }_{ 0 } }{ R } =0\Rightarrow \cfrac { F }{ m } =\cfrac { { { B }^{ 2 }d }^{ 2 }{ v }_{ 0 } }{ R } \Rightarrow { v }_{ 0 }=\cfrac { FR }{ { B }^{ 2 }{ d }^{ 2 } } $$
(c) velocity at line $$t$$
$$a=-\cfrac { dv }{ dt } \Rightarrow \int _{ 0 }^{ v }{ \cfrac { dv }{ RF-{ I }^{ 2 }{ B }^{ 2 }v }  } =\int _{ 0 }^{ v }{ \cfrac { dt }{ mR }  } ={ \left[ { I }_{ n }\left[ RF-{ I }^{ 2 }{ B }^{ 2 }v \right] \cfrac { 1 }{ -{ I }^{ 2 }{ B }^{ 2 } }  \right]  }_{ 0 }^{ v }{ \left[ \cfrac { t }{ Rm }  \right]  }_{ 0 }^{ t }\Rightarrow { I }_{ n }\left[ RF-{ I }^{ 2 }{ B }^{ 2 }v \right] -\ln { RF } =\cfrac { -{ t }^{ 2 }{ B }^{ 2 }t }{ Rm } $$
$$\Rightarrow 1-\cfrac { { I }^{ 2 }{ B }^{ 2 }v }{ RF } ={ e }^{ \cfrac { -{ I }^{ 2 }{ B }^{ 2 }t }{ Rm }  }\Rightarrow \cfrac { { I }^{ 2 }{ B }^{ 2 }v }{ RF } =1-{ e }^{ \cfrac { -{ I }^{ 2 }{ B }^{ 2 }t }{ Rm }  }\Rightarrow v=\cfrac { FR }{ { I }^{ 2 }{ B }^{ 2 } } \left( 1-{ e }^{ \cfrac { -{ I }^{ 2 }{ B }^{ 2 }t }{ Rm }  } \right) ={ v }_{ 0 }\left( 1-{ e }^{ -F{ v }_{ 0 }m } \right) $$

(a) Work done per unit test charge
$$=\phi E.dI\quad \phi E.DI=e$$
$$\Rightarrow E\phi dI=\cfrac { d\phi  }{ dt } \Rightarrow E2\pi r=\cfrac { dB }{ dt } \times A\Rightarrow E2\pi r=r{ \pi  }^{ 2 }\cfrac { dB }{ dt } \Rightarrow E=\cfrac { r{ \pi  }^{ 2 } }{ 2\pi  } \cfrac { dB }{ dt } =\cfrac { r }{ 2 } \cfrac { dB }{ dt } $$
(b) When square is considered
$$\phi EdI=e\phi E\times 2r\times 4=\cfrac { dB }{ dt } { \left( 2r \right)  }^{ 2 }\Rightarrow E=\cfrac { dB }{ dt } \cfrac { 4{ r }^{ 2 } }{ 8r } \Rightarrow E=\cfrac { r }{ 2 } \cfrac { dB }{ dt } $$
the electic field at the point $$p$$ has the same value as (a)

(a) Let $$N$$ be the number of ions that are separated by the machine per unit time. The current is $$i=qN$$ and the mass that is separated per unit time is $$M=mN$$, where $$m$$ is the mass of a single ion. $$M$$ has the value

$$M= \dfrac{100\times10^{-6}kg}{3600s} = 2.78\times10^{-8}kg/s$$.

Since, $$N=\dfrac{M}{m}$$ we have

$$i=\dfrac{qM}{m}$$

$$=\dfrac{(3.20\times10^{-19}C)(2.78\times10^{-8}kg/s)}{3.92\times10^{-25}kg}$$

$$=2.27\times10^{-2}A$$.

(b) Each ton deposits energy $$qV$$ in the cup, so the energy deposited in the time $$\Delta t$$ is given by

$$E = NqV\Delta t = iV\Delta t$$.

For $$\Delta t = 1.0h$$,

$$E = (2.27\times10^{-2}A)(100\times10^{3}V)(3600s) = 8.17\times10^{6}J$$.

To obtain the second expression, $$i/q$$ is substituted for $$N$$.

(a) We seek the electrostatic field established by the separation of charges (brought on by the magnetic force).

With $$eE=ev_{d}B$$, we define the magnitude of the electric field as

$$|\vec{E}|=v|\vec{B}|$$

$$=(20.0m/s)(0.030T)$$

$$=0.600V/m$$.

Its direction is inferred from figure; its direction is opposite to that defined by $$\vec{v} \times\vec{B}$$. In summary,

$$\vec{E} = (0.600V/m)\hat{k}$$ which insured that $$\vec{F} = q(\vec{E} + \vec{v} \times \vec{B})$$ vanishes.


(b) $$V=Ed$$

$$=(0.600V/m)(2.00m)$$

$$=1.20V$$.

For a free charge $$q$$ inside the metal strip with velocity $$\vec{v}$$ we have
$$\vec{F} = q(\vec{E} + \vec{v} \times \vec{B})$$.

We set this force equal to zero and use the relation between (uniform) electric field and potential difference. Thus,

$$v= \dfrac{E}{B}$$

$$=\dfrac{|v_{x}-v_{y}|/d_{xy}}{B}$$

$$=\dfrac{(3.90 \times 10^{-9}V )}{(1.20 \times 10^{-3}T)(0.850 \times 10^{-2}m)}$$

$$=0.382m/s$$.

Since the total force given by $$\vec{F}=(\vec{E}+\vec{v}\times \vec{B})$$ vanishes, the electric field $$\vec{E}$$ must be perpendicular to both the particle velocity $$\vec{v}$$ and the magnetic field $$\vec{B}$$. The magnetic field is perpendicular to the velocity, so $$\vec{v}\times \vec{B}$$ has magnitude $$vB$$ and the magnitude of the electric field is given by $$E=vB$$.

Since the particle has charge $$e$$ and is accelerated through a potential difference $$V$$,

$$\dfrac{mv^{2}}{2} = eV$$ and $$v=\sqrt{2\dfrac{eV}{m}}$$.
Thus,

We use $$\vec{\tau} = \vec{\mu} \times \vec{B}$$ where $$\mu$$ is the magnetic dipole moment of the wire loop and $$\vec{B}$$ is the magnetic field, as well as Newton’s second law.

Since the plane of the loop is parallel to the incline the dipole moment is normal to the incline. The forces acting on the cylinder are the force of gravity $$mg$$, acting downward from the center of mass, the normal force of the incline $$F_{N}$$, acting perpendicularly to the incline through the center of mass, and the force of friction $$f$$, acting up the incline at the point of contact. We take the $$x$$ axis to be positive down the incline. Then the $$x$$ component of Newton’s second law for the center of mass yields

$$mg sin \theta - f = ma$$.

For purposes of calculating the torque, we take the axis of the cylinder to be the axis of rotation. The magnetic field produces a torque with magnitude $$μB sin \theta$$, and the force of friction produces a torque with magnitude $$fr$$, where $$r$$ is the radius of the cylinder. The first tends to produce an angular acceleration in the counterclockwise direction, and the second tends to produce an angular acceleration in the clockwise direction. Newton’s second law for rotation about the center of the cylinder, $$\tau = I \alpha$$, gives

$$fr - \mu B sin \theta = I \alpha$$.

Since we want the current that holds the cylinder in place, we set $$a = 0$$ and $$\alpha = 0$$, and use one equation to eliminate $$f$$ from the other. The result is $$mgr = \mu B$$. The loop is rectangular with two sides of length $$L$$ and two of length $$2r$$, so its area is $$A = 2rL$$ and the dipole moment is $$\mu = NiA = Ni(2rL)$$. Thus, $$mgr = 2NirLB$$ and

$$i = \dfrac{mg}{2NLB}$$

$$=\dfrac{(0.250kg)(9.8m/s^{2})}{2(10.0)(0.100m)(0.500T)}$$

$$=2.45A$$.

Let $$v_{||} = v cos{\theta}$$.

The electron will proceed with a uniform speed $$v_{1}$$ in the direction of $$\vec{B}$$ while undergoing uniform circular motion with frequency $$f$$ in the direction perpendicular to $$B$$:

$$f = \dfrac{eB}{2 \pi m_{e}}$$.

The distance $$d$$ is then

$$ d = v_{||}T$$

$$=\dfrac{v_{||}}{f}$$

$$=\dfrac{(vcos\theta)2 \pi m_{e}}{eB}$$

$$=\dfrac{2\pi(1.5 \times 10^{7}m/s )(9.11 \times 10^{-31}kg)(cos10^{0})}{(1.60 \times 10^{-19}C)(1.0 \times 10^{-3}T)}$$

$$=0.53m$$.

When the spring is streatched and released, the wire $$AB$$ will excute simple harmonic motion, so induced emf will vary periodically. At $$t=0$$, wire is at the extreme position
$$v=0$$
$$v=A \omega \sin \omega t$$
Induced emf $$ \varepsilon =Bvl$$
$$=BA \omega \sin \omega t$$
Where $$A=BB'=AA'$$ is the amplitude of motion and $$\omega$$ is angular frequency.

$$\cdot$$ Direction changes.
$$\cdot$$ Magnitude changes.

We write the equation of the circuit as,
$$Ri = \dfrac {L}{\eta} \dfrac {di}{dt} = \xi$$,
for $$t\geq 0$$. The current at $$t = 0$$ just after inductance is changed, is
$$i = \eta \dfrac {\xi}{R}$$, so that the flux through the inductance is unchanged.
We look for a solution in the form
$$i = A + Be^{-t/C}$$
Substituting $$C = \dfrac {L}{\eta R}, B = \eta - 1, A = \dfrac {\xi}{R}$$
Thus, $$i = \dfrac {\xi}{R} (1 + (\eta - 1)e^{-\eta Rt/ L})$$.