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