Faraday's Law
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A long solenoid of radius 2 cm has 100 turns/cm and carries a current of 5 A. A coil of radius 1 cm having 100 turns and a total resistance of 20ω is placed inside the solenoid coaxially. The coil is connected to a galvanometer. If the current in the solenoid is reversed in direction, find the charge flown through the galvanometer.
- Electric field lines are parallel to the surface
- Electric field lines are perpendicular to the surface
- Electric field lines are at 60∘ to the surface
- Electric field lines are at 45∘ to the surface.
An electric dipole is held in a uniform electric field. Show that the net force acting on it is zero. The dipole is aligned parallel to the field. Find the work done in rotating it through the angle of 180 degrees.
A copper disc of radius 0.1 m rotates about its centre with 10 revolutions per second in a uniform magnetic field of 0.1 Tesla. The emf induced across the radius of the disc is
- π10 V
- 2π10 V
- 10 π m V
- 20 π m V
- In gravity free space, a positively charged bead cannot be in equilibrium at any position.
- In gravity free space, a negatively charged bead cannot be in equilibrium at any position.
- In the presence of gravity, a negatively charged bead cannot be in equilibrium position above the plane of fixed charges.
- In the presence of gravity, a positively charged particle can be in equilibrium above the plane of the fixed charges.
Given: the linear density of wire is 46.6 g/m
- 1.6×10−2 T
- 2.48×10−3 T
- 3.2×10−2 T
- 4.2×10−2 T
Given sin37∘=0.6, cos37∘=0.8
(i) The surface(s) that have zero electric flux through them are
[1 Mark]
- S1 and S3
- S5 and S6
- S2 and S4
- S1 and S5
- 1.23
- 1.63
- 1.83
- 1.51
[0.77 Mark]
- 28×10−10 Nm
- 8×10−9 Nm
- 81×10−6 Nm
- 2×10−4 Nm
- The two coils show attraction.
- The two coils show repulsion.
- There is no change in the position of two coils.
- Induction of current is not possible in coil B.
The above configuration shows three identical bulbs, grade them in order of their brightness.
- B3>B2=B1
- B1=B2>B3
- B3>B1>B2
- B2>B1>B3
- Double
- Three times
- Four times
- One - fourth
- 12B0IL^j
- 12B0IL^i
- B0IL^j
- BoIL^i
The half circle lies in a uniform changing magnetic field B=4t2+2t+5 (SI unit), where t is the time in
seconds.
An ideal battery E = 2 V is connected as shown and the total resistance of the wire is 2Ω. The net current in the loop at t = 5 seconds is:
- 1 A
- 1.5 A
- 0.10 A
- 0.26 A
- BIL
- BIL2
- BIL2
- BIL22
- 5 mA
- 10 mA
- 15 mA
- 20 mA
- V02
- 0
- V0√2
- V0√3
- BR2
- 2BR
- BR4
- BR
- Net force on wire is zero
- Net magnetic force on wire is BIR√2
- Tension in the threads is BIR
- Tension in the threads is BIR√2
- 4
- 6
- 7
- 8
A conducting square loop having edges of length 2.0 cm is rotated through 180∘ about a diagonal in 0.20 a. A magnetic field B exists in the region which is perpendicular to the loop in its initial position. If the average induced emf during the rotation is 20 mV, find the magnitude of the magnetic field.
- Electric field lines are parallel to the surface
- Electric field lines are perpendicular to the surface
- Electric field lines are at 60∘ to the surface
- Electric field lines are at 45∘ to the surface.
- 1.5 ms
- 2.5 ms
- 3.5 ms
- 4.5 ms