Q 1 :    

A ceiling fan having 3 blades of length 80 cm each is rotating with an angular velocity of 1200 rpm. The magnetic field of earth in that region is 0.5 G and the angle of dip is 30°. The emf induced across the blades is Nπ×10-5 V. The value of N is ______.                            [2024]



(32)

 



Q 2 :    

A horizontal straight wire 5 m long extending from east to west falling freely at right angle to horizontal component of earth's magnetic field 0.60×10-4 Wbm-2. The instantaneous value of emf induced in the wire when its velocity is 10ms-1 is _____ ×10-3 V.                           [2024]



(3)          BH=0.60×10-4Wb/m2

              ε=BHvl=0.60×10-4×10×5

                 =3×10-3V

 



Q 3 :    

A rod of length 60 cm rotates with a uniform angular velocity 20 rad s-1 about its perpendicular bisector, in a uniform magnetic field 0.5 T. The direction of magnetic field is parallel to the axis of rotation. The potential difference between the two ends of the rod is ______ V.     [2024]



(0)

VB-VC=B0ωl28

VA-VC=B0ωl28

VB-VA=0

 



Q 4 :    

A square loop PQRS having 10 turns, area 3.6×10-3m2 and resistance 100 Ω is slowly and uniformly being pulled out of a uniform magnetic field of magnitude B = 0.5 T as shown. Work done in pulling the loop out of the field in 1.0 s is ___ ×10-6J.              [2024]



(3)

W=heat loss

=Δϕ2RΔt=(3.6×10-3×0.5×10)2100×1

=182100×10-6=3.24×10-6 J



Q 5 :    

A rectangular loop of sides 12 cm and 5 cm, with its sides parallel to the x-axis and y-axis respectively, moves with a velocity of 5 cm/s in the positive x-axis direction, in a space containing a variable magnetic field in the positive z-direction. The field has a gradient of 10-3 T/cm along the negative x-direction and it is decreasing with time at the rate of 10-3 T/s. If the resistance of the loop is 6 , the power dissipated by the loop as heat is ______ ×10-9 W.        [2024]



(216)

B0 is the magnetic field at origin

dBdx=-10-310-2B0BdB=-0x10-1dx

B-B0=-10-1xB=(B0-x10)

Motional emf in AB=0

Motional emf in CD=0

Motional emf in AD=ε1=B0lv

Magnetic field on rod BC,

B=(B0-(-12×10-2)10)

Motional emf in BC=ε2=(B0+12×10-210)l×v

εeq=ε2-ε1=300×10-7 V

For time variation

(εeq)'=AdBdt=60×10-7 V

(εeq)net=εeq+(εeq)'=360×10-7 V

Power=(εeq)net2R=216×10-9 W



Q 6 :    

A uniform magnetic field of 0.4 T acts perpendicular to a circular copper disc 20 cm in radius. The disc is having a uniform angular velocity of 10π rads1 about an axis through its centre and perpendicular to the disc. What is the potential difference developed between the axis of the disc and the rim? (π = 3.14)          [2025]

  • 0.0628 V

     

  • 0.5024 V

     

  • 0.2512 V

     

  • 0.1256 V

     

(3)

Given : B = 0.4 T, r = 20 cm and ω=10π rad/s

The potential difference developed between the axis of the disc and the rim, e=12Bωr2 = 0.2512 V



Q 7 :    

Figure

A conducting bar moves on two conducting rails as shown in the figure. A constant magnetic field B exists into the page. The bar starts to move from the vertex at time t = 0 with a constant velocity. If the induced EMF is Etn, then value of n is ______.          [2025]



1

Figure

E=lvB

E=2x3×vB and x=vt

E=23v2Bt     Et1



Q 8 :    

Conductor wire ABCDE with each arm 10 cm in length is placed in magnetic field of 12 Tesla, perpendicular to its plane. When conductor is pulled towards right with constant velocity of 10 cm/s, induced emf between points A and E is ______ mV.          [2025]

Figure



10

As field is uniform, we can replace the bent wire with straight wire from A to E. EMF ε=Bvl

ε=12×10100×2(10 sin 45°)100=0.01 V=10 mV