हिंदी

Use the Expression for Lorentz Force Acting on the Charge Carriers of a Conductor to Obtain the Expression for the Induced Emf Across the Conductor of Length L

Advertisements
Advertisements

प्रश्न

Use the expression for Lorentz force acting on the charge carriers of a conductor to obtain the expression for the induced emf across the conductor of length l moving with velocity v through a magnetic field B acting perpendicular to its length.

संक्षेप में उत्तर
Advertisements

उत्तर

The arm PQ is moved to the left side, thus decreasing the area of the rectangular loop. This movement induces a current I as shown.

Let us consider a straight conductor moving in a uniform and time-independent magnetic field. The figure shows a rectangular conductor PQRS in which the conductor PQ is free to move. The rod PQ is moved towards the left with a constant velocity v as shown in the figure. Assume that there is no loss of energy due to friction. PQRS forms a closed circuit enclosing an area that changes as PQ moves. It is placed in a uniform magnetic field B which is perpendicular to the plane of this system. If the length RQ = x and RS l the magnetic flux ∅B enclosed by the loop PQRS will be
B = Blx

Since x is changing with time, the rate of change of flux φB will induce an emf given by: 

`ε = -(d∅_B)/(dt) = - (d(Bl"x"))/(dt)`

= `-"B"l (d"x")/(d"t") = "B"l"v"`

where we have used dx/dt = – v which is the speed of the conductor PQ. The induced emf `"B"lv"` called motional emf. Thus, we are able to produce induced emf by moving a conductor instead of varying the magnetic field, that is, by changing the magnetic flux enclosed by the circuit. It is also possible to explain the motional emf expression by invoking the Lorentz force acting on the free charge carriers of conductor PQ. Consider any arbitrary charge q in the conductor PQ. When the rod moves with speed v, the charge will also be moving with speed v in the magnetic field B. The Lorentz force on this charge is qvB in magnitude, and its direction is towards Q. All charges experience the same force, in magnitude and direction, irrespective of their position in the rod PQ.

The work done in moving the charge from P to Q is,

`W = q"vB"l`

Since emf is the work done per unit charge,

`epsilon = "W"/q`

= `"B"l"v"`

This equation gives emf induced across the rod PQ
The total force on the charge at P is given by

`vec"F"= q(vec"E" + vec"v" xx vec"B")`

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
2014-2015 (March) Ajmer Set 2

वीडियो ट्यूटोरियलVIEW ALL [2]

संबंधित प्रश्न

Find the value of t/τ for which the current in an LR circuit builds up to (a) 90%, (b) 99% and (c) 99.9% of the steady-state value.


In a series, LCR circuit, obtain an expression for the resonant frequency.


Answer the following question.
In a series LCR circuit connected across an ac source of variable frequency, obtain the expression for its impedance and draw a plot showing its variation with frequency of the ac source. 


Answer the following question.
What is the phase difference between the voltages across the inductor and the capacitor at resonance in the LCR circuit? 


Choose the correct answer from given options
The phase difference between the current and the voltage in series LCR circuit at resonance is


At resonance frequency the impedance in series LCR circuit is ______.


At resonant frequency the current amplitude in series LCR circuit is ______.


A series LCR circuit containing a 5.0 H inductor, 80 µF capacitors, and 40 Ω resistor is connected to a 230 V variable frequency ac source. The angular frequencies of the source at which power is transferred to the circuit are half the power at the resonant angular frequency are likely to be ______.


A series LCR circuit containing 5.0 H inductor, 80 µF capacitor and 40 Ω resistor is connected to 230 V variable frequency ac source. The angular frequencies of the source at which power transferred to the circuit is half the power at the resonant angular frequency are likely to be ______.


In series LCR circuit, the plot of Imax vs ω is shown in figure. Find the bandwidth and mark in the figure.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×