1. A rectangular, a square, a circular and an elliptical loop, all in the (x – y) plane, are moving out of a uniform magnetic field with a constant velocity,
   →
   V = vî.
   The magnetic field is directed along the negative z axis direction. The induced emf, during the passage of these loops, out of the field region, will not remain constant for

1. 1. the circular and the elliptical loops.

   
   
   

   2. only the elliptical loop. ​

   
   
   

   3. any of the four loops.

   
   
   

   4. the rectangular, circular and elliptical loops.

   
   
   

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1. View Hint View Answer Discuss in Forum

   The induced emf will remain constant only in the case of rectangular and square loops. In case of the circular and the elliptical loops, the rate of change of area of the loops during their passage out of the field is not constant, hence induced emf will not remain constant for them.

   Correct Option: A

   The induced emf will remain constant only in the case of rectangular and square loops. In case of the circular and the elliptical loops, the rate of change of area of the loops during their passage out of the field is not constant, hence induced emf will not remain constant for them.

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2. A circular disc of radius 0.2 meter is placed in a uniform magnetic field of induction 1/π (Wb/m^2→_B .The magnetic flux linked with the disc is:​​

1. 1. 0.02 Wb ​

   
   
   

   2. ​0.06 Wb ​

   
   
   

   3. 0.08 Wb

   
   
   

   4. 0.01 Wb

   
   
   

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1. View Hint View Answer Discuss in Forum

   Here,
   

   B =	1	(Wb/m2)
   	π

   
   θ = 60°
   Area normal to the plane of the disc
   

   = πr2cos60° =	πr2
   	2

   
   Flux = B × normal area
   

   =	0.2 × 0.2	= 0.02 Wb
   	2

   Correct Option: A

   Here,
   

   B =	1	(Wb/m2)
   	π

   
   θ = 60°
   Area normal to the plane of the disc
   

   = πr2cos60° =	πr2
   	2

   
   Flux = B × normal area
   

   =	0.2 × 0.2	= 0.02 Wb
   	2

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3. A conducting circular loop is placed in a uniform magnetic field of 0.04 T with its plane perpendicular to the magnetic field. The radius of the loop starts shrinking at 2 mm/s. The induced emf in the loop when the radius is 2 cm is

1. 1. 4.8 πμ V ​

   
   
   

   2. 0.8 πμ V ​

   
   
   

   3. 1.6 πμ V ​

   
   
   

   4. 3.2 πμ V

   
   
   

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1. View Hint View Answer Discuss in Forum

   ​Induced emf in the loop is given by
   

   e = - B.	dA
   	dt

   
   where A is the area of the loop.
   

   e = - B.	d	(π r2) = - Bπ2r	dr
   	dt		dt

   
   r = 2cm = 2 × 10^–2 m ​
   dr = 2 mm = 2 × 10^–3
   m ​dt = 1s
   

   e = - 0.04 × 3.14 × 2 × 2 × 10-2 ×	2 × 10-3	V
   	1

   
   = 0.32 π × 10^–5 V
   = 3.2 π ×10^–6 V
   ​= 3.2 π μV

   Correct Option: D

   ​Induced emf in the loop is given by
   

   e = - B.	dA
   	dt

   
   where A is the area of the loop.
   

   e = - B.	d	(π r2) = - Bπ2r	dr
   	dt		dt

   
   r = 2cm = 2 × 10^–2 m ​
   dr = 2 mm = 2 × 10^–3
   m ​dt = 1s
   

   e = - 0.04 × 3.14 × 2 × 2 × 10-2 ×	2 × 10-3	V
   	1

   
   = 0.32 π × 10^–5 V
   = 3.2 π ×10^–6 V
   ​= 3.2 π μV

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4. As a result of change in the magnetic flux linked to the closed loop shown in the Fig,
   
   an e.m.f. V volt is induced in the loop. The work done (joules) in taking a charge Q coulomb once along the loop is

1. 1. QV ​

   
   
   

   2. 2QV

   
   
   

   3. ​QV/2

   
   
   

   4. Zero

   
   
   

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1. View Hint View Answer Discuss in Forum

   ξ =	W	⇒ V =	W	⇒ W = QV
   	Q		Q

   Correct Option: A

   ξ =	W	⇒ V =	W	⇒ W = QV
   	Q		Q

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5. A conducting circular loop is placed in a uniform magnetic field, B = 0.025 T with its plane perpendicular to the loop.​​
   ​The radius of the loop is made to shrink at a constant rate of 1 mm s^–1. The induced e.m.f. when the radius is 2 cm, is

1. 1. 2πμ V

   
   
   

   2. πμ V

   
   
   

   3. 	π	μ V
      	2

   
   
   

   4. 2μ V

   
   
   

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1. View Hint View Answer Discuss in Forum

   Magnetic flux linked with the loop is
   φ = Bπr²
   

   |e| =	dφ	= B π .2r	dr
   	dt		dt

   
   When r = 2 cm,
   

   	dr	= 1 mm s-1
   	dt

   
   e = 0.025 × π × 2 × 2 × 10^–2 × 10^–3
   = 0.100 × π × 10^–5 = π × 10^–6 V = π μ V

   Correct Option: B

   Magnetic flux linked with the loop is
   φ = Bπr²
   

   |e| =	dφ	= B π .2r	dr
   	dt		dt

   
   When r = 2 cm,
   

   	dr	= 1 mm s-1
   	dt

   
   e = 0.025 × π × 2 × 2 × 10^–2 × 10^–3
   = 0.100 × π × 10^–5 = π × 10^–6 V = π μ V

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