Name: Magnetic Flux (Part-2)
Uploaded: 2019-07-19
Duration: 6 min 14 s
Description: This video enables students to understand magnetic flux for closed objects and curved surfaces placed in the non-uniform magnetic field. Click to learn more.

Question

Show that the induced emf produced by changing the area of a rectangular coil placed perpendicular to a uniform magnetic field is

- B l v

, where the symbols have their usual meanings.

Accepted Answer

It is evident from Faraday’s experiments that whenever there is a change in the magnetic flux passing through a closed circuit, an electric current is induced in the circuit. It can be explained on the basis of Lorentz force.

Suppose a conducting-rod JK (Fig. $8$ ) is being moved without friction on the arms of a U-shaped stationary conductor $MNOL$ with a velocity $\vec{v}$ towards right. The conductor is situated in a magnetic field $\vec{B}$ perpendicular to the plane of paper directed downwards. Due to the motion of the conducting-rod, the free electrons present in the rod are acted upon by a magnetic (Lorentz) force $\vec{F_{m}}$ of magnitude $q v B$ which takes the electrons from the end $J$ of the rod to the end $K$ . Since, a closed circuit is available to the electrons, they drift along the path $J \to K \to O \to N \to J$ . Thus, an electric current is established in the circuit along $J \to N \to O \to K \to J$ (anticlockwise)*. So long the rod $J K$ is kept moving in the magnetic field, the electric current continues to flow in the circuit. It means that an emf is induced in the moving rod which maintains the current in the circuit. Since, this emf induced in the rod is due to the motion of the rod, it is also called as the ‘motional emf. It is due to the Lorentz forces acting on the free electrons in the moving rod.
Suppose, the induced emf in the moving rod $J K$ is $e$ and the induced current in the circuit is $I$ . We know that when a current-carrying conductor is in a magnetic field, it is acted upon by a force in a direction given by the Fleming’s left-hand rule. The magnitude of the force imposed by the magnetic field $\vec{B}$ on the current-carrying rod $J K$ is given by

$F = I l B,$

where $l$ is the length of the rod $J K$ . The force $F$ is directed towards left. Hence, to keep the rod moving with a constant velocity $v$ towards right, a force $F_{e x t}$ equal and opposite to $F$ will have to be applied on it; that is,

$F_{e x t} = - F = - I l B$

Suppose, under the force $F_{e x t}$ , the rod $J K$ undergoes a small displacement $∆ \vec{x}$ in the direction of the force in a time-interval $∆ t$ and comes in the position $J' K'$ . Then, the work done on the rod will be given by

$W = F_{e x t} Δ x = - I l B Δ x$

But $Δ x = v \times Δ t (magnitude of velocity \times time - interval).$

$∴ W = - I l B v Δ t$ .

But $I \times Δ t = q (charge flowing through the circuit in time Δ t)$

$∴ W = - B v l q .$

This work provides the necessary energy for the flow of charge in the circuit. We know that the energy supplied by a cell in flowing unit charge through a circuit is called the emf of the cell. Hence, the induced emf e in the rod $J K$ (which is working as a cell in the circuit $J N O K J$ ) is given by

${ΔΦ}_{B} = magnitude of magnetic field \times change in area ({JJ}^{'} K^{'} K) = B \times (l \times Δ x)$

Hence, the rate of change of magnetic flux is

$\frac{{ΔΦ}_{B}}{Δt} = B l \frac{Δx}{Δt}$ .

But $Δ x / Δ t = v$ (velocity of the rod). Thus,

$\frac{{ΔΦ}_{B}}{Δ t} = B l v . . . . (ii)$

Comparing Eqs. (i) and (ii), we get

$e = - \frac{{ΔΦ}_{B}}{Δ t}$ .

In the limit $∆ t \to 0,$

$e = - \frac{{dΦ}_{B}}{d t} .$

This is Faraday’s law of electromagnetic induction.
The negative sign signifies Lenz’s law. The direction of current (anticlockwise) in the circuit due to the induced emf $e$ is such that the force imposed on the rod (due to the current) opposes the motion of the rod (the motion of the rod is the cause of the current).

Now, let $R$ be the resistance of the conducting rod $J K$ . The resistance of $U -$ shaped stationary conductor $M N O L$ is negligible compared to $R$ and the resistance of rod $J K$ , that is, $R$ does not change as it moved. So the current $I$ in the loop $J K O N$ is,

$I = \frac{e}{R} = \frac{B l v}{R} . . . (iii)$

Further, due to presence of magnetic field, force on the conductor $J K$ (opposite to $\vec{v}$ ) will be

$F = I I B = \frac{B^{2} l^{2} v}{R}$

Now, for pushing this conductor in the magnetic field with a constant velocity $v$ , the power required

$P = F v = \frac{B^{2} l^{2} v^{2}}{R} . . . (iv)$

The work done by the external agent to move the rod with uniform velocity $v$ in the magnetic field is mechanical in nature, part of the work is dissipated as Joule’s heat and remaining part is converted into the electrical energy. Power loss due to Joule’s heating,

$P_{H} = I^{2} R = {(\frac{B l v}{R})}^{2} \times R = \frac{B^{2} l^{2} v^{2}}{R} .$

Dimensions of Induced EMF: Numerically, we have

$| e | = \frac{{dΦ}_{B}}{d t} (where | e | is the magnitude of' e' only)$

$∴ Dimensions of e = \frac{Dimension of Φ_{B}}{Dimension of t} = \frac{[{ML}^{2} T^{- 2} A^{- 1}]}{[T]} = [{ML}^{2} T^{- 3} A^{- 1}]$

Show that the induced emf produced by changing the area of a rectangular coil placed perpendicular to a uniform magnetic field is $- B l v$ , where the symbols have their usual meanings.

Important Questions on Electromagnetic Induction

Learn and Prepare for any exam you want !

Show that the induced emf produced by changing the area of a rectangular coil placed perpendicular to a uniform magnetic field is - B l v, where the symbols have their usual meanings.

Important Questions on Electromagnetic Induction

Learn and Prepare for any exam you want !

Show that the induced emf produced by changing the area of a rectangular coil placed perpendicular to a uniform magnetic field is $- B l v$ , where the symbols have their usual meanings.