The following figure shows two copper rods moving with the same velocity parallel to a long straight wire carrying current. If the induced emf in rod AB is \(\mathrm{E}\), the Induced emf in rod CD is:

           
1. \(\mathrm{E}\)
2. \(\mathrm{E}\cos \theta\)
3. \(\mathrm{E}\sin\theta\)
4. \(0\)

Subtopic:  Motional emf |
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A triangular wire frame, in the form of an equilateral triangle \(PQR\) moves with a uniform velocity into a region where there is a uniform magnetic field \(B\). The edge \(PQ\) is parallel to the boundary of the region and the velocity \(v\) is perpendicular to it. The emf(\(E\)) induced within the frame is plotted as a function of time \(t,\) starting from when the frame enters the magnetic field. \(E\) is given by:
1. \(Bv^2t\) 2. \(2Bv^2t\)
3. \(\dfrac{\sqrt3}{2}Bv^2t\) 4. \(\dfrac{2}{\sqrt3}Bv^2t\)
Subtopic:  Motional emf |
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A circular wire of radius \(R\) is placed in a uniform magnetic field \(B,\) which acts into the plane as shown. The wire is given a half-turn about a diameter. The resistance per unit length of the wire is \(\lambda.\) The total charge flowing through the wire is:
                              

 
1. \(\dfrac{2BR}{\lambda}\) 2. \(\dfrac{BR}{\lambda}\)
3. \(\dfrac{BR}{2\lambda}\) 4. zero
Subtopic:  Motional emf |
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The current through the inductor in the figure is initially zero. The initial rate of change of the current \(i\) through the inductor (i.e. \(\dfrac{di}{dt}\)) is:
           
 
1. zero 2. \(-\dfrac{I_{0} R}{L}\)
3. \(\dfrac{I_{0} R}{L}\) 4. \(\dfrac{I_{0} R}{2L}\)
Subtopic:  LR circuit |
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Suppose a long rectangular loop of width \(w\) is moving along the \(x\)-direction with its left arm in a magnetic field perpendicular to the plane of the loop (see figure). The resistance of the loop is zero and it has an inductance \(L.\) At time, \(t= 0,\) its left arm passes the origin, \(O.\)
       
If for \(t\geq0,\) the current in the loop is \(I\) and the distance of its left arm from the origin is \(x,\) then \(I\) versus \(x\) graph would be:
 
1. 2.
3. 4.
Subtopic:  Motional emf |
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A conducting rod of mass \(m\) and length \(l\) is free to move without friction on two parallel long conducting rails, as shown below. There is a resistance \(R\) across the rails and in the entire space around there is a uniform magnetic field \(B\) normal to the plane of the rod and the rails. The rod is given an impulsive velocity \(v_0.\)

            

Finally, the initial energy \(\dfrac{1}{2}mv^2_0,\)

1. will be converted fully into heat energy in the resistor.
2. will enable the rod to continue to move with velocity \(v_2\) since the rails are frictionless.
3. will be converted fully into magnetic energy due to induced current.
4. will be converted into the work done against the magnetic field.
Subtopic:  Motional emf |
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A parallel plate capacitor is being charged by means of a constant current \(i.\) The plates are circular (of radius \(R)\) and are separated by a distance \(d.\) The magnetic field between the plates, at a distance \(\dfrac{R}{2}\) from the central axis is:
1. zero 2. \(\dfrac{\mu_{0} i}{2 \pi \left(\dfrac{R}{2}\right)}\)
3. \(\dfrac{1}{4}\dfrac{\mu_{0} i}{2 \pi R}\) 4. \(\dfrac{1}{2}\dfrac{\mu_0 i}{2\pi R}\)
Subtopic:  Magnetic Flux |
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The two long, parallel wires shown in the diagram carry equal and opposite currents \(i\). The currents change linearly with time: \(\dfrac{di} {dt}\) = a constant = \(K\). The small circuit is situated midway between the wires and has an area \(A\). The emf induced in the small circuit is: 
1. zero 2. \(\dfrac{\mu_{0} A K}{2 \pi l}\)
3. \(\dfrac{\mu_{0} A K}{ \pi l}\) 4. \(\dfrac{2 \mu_{0} A K}{\pi l}\)
Subtopic:  Magnetic Flux |
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Rings are rotated and translated in a uniform magnetic field as shown in the figure. Arrange the magnitude of emf induced across \(AB\)

         

1. \(\mathrm{emf_{a}<emf_{b}<emf_{c}}\)
2. \(\mathrm{emf_{a}=emf_{b}<emf_{c}}\)
3. \(\mathrm{emf_{a}={emf}_{c}<{emf}_{b}}\)
4. \(\mathrm{emf_{a}<emf_{b}={emf}_{c}}\)
Subtopic:  Motional emf |
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The network shown in the figure is part of a complete circuit. If the current flowing in the circuit is \(I=(10t+5)~\text A,\) the potential difference \((V_B-V_A)\) at \(t=0\) is:

     
1. \(15~\text V\) 2. \(5~\text V\)
3. \(-5~\text V\) 4. \(-15~\text V\)
Subtopic:  Self - Inductance |
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