Electromagnetic Induction Test

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Electromagnetic Induction Test - 8

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Question 1
1 / -0.25

The magnetic field in a region is given by . A square loop of edge-length d is placed with its edge along x &y axis. The loop is moved with constant velocity . The emf induced in the loop is

A

B

C

D
None

Solution

By faraday 's law
ϵ=−d ϕ/dt =d ϕdx /dx dt
=B_o d (2d /2a)v_o
Question 2
1 / -0.25

Consider the situation shown in fig. The resistanceless wire AB is slid on the fixed rails with a constant velocity. If the wire AB is replaced by a resistanceless semicircular wire, the magnitude of the induced current will

A
increase

B
remain the same

C
decrease

D
increase or decrease depending on whether the semicircle bulges towards the resistance or away from it

Solution

Rate of change of area with remain same even if wire is replaced by semi-circular wire,
∴induced current will be same as earlier one as I = (e/R) &e i.e. induced emf is function of area because
e = –N (d ɸ/dt) = –N (d/dt) (ABcoswt).
Question 3
1 / -0.25

A conducting square loop of side I and resistance R moves in its plane with a uniform velocity v perpendicular to one of its sides. A uniform and constant magnetic field B exists along the perpendicular to the plane of the loop in fig. The current induced in the loop is

A
RvB

B
zero

C
vBL/R

D
vBL

Solution

Voltage is induced due to change in the magnetic flux through a closed loop.
Here
Magnetic Flux = ϕ=BA=const.
Hence
Induced EMF = dtd ϕ=0
Induced current =i=E/R=0
Question 4
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Two identical conductors P and Q are placed on two frictionless fixed conducting rails R and S in a uniform magnetic field directed into the plane. If P is moved in the direction shown in figure with a constant speed, then rod Q

A
will be attracted towards P

B
will be repelled away from P

C
will remain stationary

D
may be repelled or attracted towards P

Solution

As the conductor P moves away from Q, the area of the loop enclosed by the conductors and the rails increases. This in turn increases the flux through the loop.
EMF will be induced in such a way that the change in flux will be resisted. The induced current will cause Q to move towards P thereby reducing the area and thus the flux back.
Question 5
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Two infinitely long conducting parallel rails are connected through a capacitor C as shown in the figure. A conductor of length l is moved with constant speed v₀ . Which of the following graph truly depicts the variation of current through the conductor with time ?

A

B

C

D

Solution

By Faraday 's Law of induction,
ε=−d ϕ/dt
=−Bl (dx/dt) =−Blv₀
This emf should induce the movement of charges creating a current. But due to the attached capacitor, all charges are conserved.
Thus I= dq/dt =0
The correct option is C.
Question 6
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A thin wire of length 2m is perpendicular to the xy plane. It is moved with velocity through a region of magnetic induction Wb/m² . Then potential difference induced between the ends of the wire

A
2 volts

B
4 volts

C
0 volts

D
None of these

Solution
Question 7
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A long metal bar of 30 cm length is aligned along a north south line and moves eastward at a speed of 10 ms^-1 . A uniform magnetic field of 4.0 T points vertically downwards. If the south end of the bar has a potential of 0V, the induced potential at the north end of the bar is

A
+ 12 V

B
-12 V

C
0 V

D
Cannot be determined since there is not closed circuit

Solution

Induced emf =Blv=12V. It is induced in the northward direction by right hand rule (emf= )
therefore if the south end of the pole has potential of 0V, the north end will have a potential of 12V.
Question 8
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A conducting rod moves with constant velocity v perpendicular to the long, straight wire carrying a current I as shown compute that the emf generated between the ends of the rod.

A

B

C

D

Solution

The magnetic field near the conducting rod due to long current carrying wire is
B= μ₀ I/2 πr (using Ampere 's law, ∫B.dl=μ₀ I)
Now, induced emf due to moving rod is e=Blv= (μ₀ I/2 πr) ×lv= μ₀ vIl/2 πr
Question 9
1 / -0.25

A square loop of side a and resistance R is moved in the region of uniform magnetic field B (loop remaining completely insidefield), with a velocity v through a distance x. The work done is

A

B

C

D
None

Solution
Question 10
1 / -0.25

A rod closing the circuit shown in figure moves along a U shaped wire at a constant speed v under the action of the force F. The circuit is in a uniform magnetic field perpendicular to the plane. Calculate F if the rate of heat generation in the circuit is Q.

A
F = Qv

B
F = Q/v

C
F = v/Q

D