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## ELECTROMAGNETIC INDUCTION Quiz-11

As per analysis for previous years, it has been observed that students preparing for NEET find Physics out of all the sections to be complex to handle and the majority of them are not able to comprehend the reason behind it. This problem arises especially because these aspirants appearing for the examination are more inclined to have a keen interest in Biology due to their medical background. Furthermore, sections such as Physics are dominantly based on theories, laws, numerical in comparison to a section of Biology which is more of fact-based, life sciences, and includes substantial explanations. By using the table given below, you easily and directly access to the topics and respective links of MCQs. Moreover, to make learning smooth and efficient, all the questions come with their supportive solutions to make utilization of time even more productive. Students will be covered for all their studies as the topics are available from basics to even the most advanced. .

Q1. A horizontal straight wire 10 m long extending from east to west is falling with a speed of 5.0 ms-1, at right angles to the horizontal component of the earth’s magnetic field of strength 0.30×10-4 Wbm-2. the instantaneous value of the induced potential gradient in the wire, from west to east, is
•  +1.5×10-3 Vm-1
•  -1.5×10-3 Vm-1
•  +1.5×10-4 Vm-1
•  -1.5×10-4 Vm-1
Solution
Given, B=0.30×10-4 Wbm-2,l=10 m and v=5.0 ms-1 The induced potential gradient V=Bvl V=-0.30×10-4×5×10 V= -1.5×10-3 Vm-1 From west to east, V=+1.5×10-3 Vm-1

Q2. The magnetic flux linked with a vector area A ⃗ in a uniform magnetic field B ⃗ is
•  B ⃗×A ⃗
•  AB
•  B ⃗∙A ⃗
•  B/A
Solution
It is a fact.

Q3. The magnetic induction in the region between the pole faces of an electromagnet is 0.7weber/m2. The induced e.m.f. in a straight conductor 10 cm long, perpendicular to B and moving perpendicular both to magnetic induction and its own length with a velocity2 m/sec is
•  0.08 V
•  0.14 V
•  0.35 V
•  0.07 V
Solution
e=Bvl⇒e=0.7×2×(10×10-2)=0.14 V

Q4. The self inductance of a solenoid of length L, area of cross-section A and having N turns is
•  (Î¼0 N2A)/L
•  (Î¼0 NA)/L
•  Î¼0 N2 LA
•  Î¼0 NAL
Solution
It is a fact.

Q5. When a bar magnet falls through a long hollow metal cylinder fixed with its axis vertical, the final acceleration of the magnet is
•  Equal to zero
•  Less than g
•  Equal to g
•  Equal to g in the beginning and then more than g
Solution
If bar magnet is falling vertically through the hollow region of long vertical copper tube then the magnetic flux linked with the copper tube (due to ‘non-uniform’ magnetic field of magnet) changes and eddy currents are generated in the body of the tube by Lenz’s law. The eddy currents oppose the falling of the magnet which therefore experience a retarding force. The retarding force increases with increasing velocity of the magnet and finally equals the weight of the magnet. The magnet then attains a constant final terminal velocity i.e., magnet ultimately falls with zero acceleration in the tube

Q6. A conducting ring is placed around the core of an electromagnet as shown in fig. when key K is pressed, the ring

•  Remain stationary
•  Is attracted towards the electromagnet
•  Jumps out of the core
•  None of the above
Solution
When key k is pressed, current through the electromagnet start increasing, i.e., flux linked with ring increases which produces repulsion effect

Q7. When a rod of length l is rotated with angular velocity of Ï‰ in a perpendicular field of induction B, about one end, the emf across its ends is
•  Bl2 Ï‰
•  (Bl2Ï‰)/2
•  BlÏ‰
•  BlÏ‰/2
Solution
A conducting rod of length l whose one end is fixed, is rotated about the axis passing through its fixed end and perpendicular to its length with constant angular velocityÏ‰. Magnetic field (B) is perpendicular to the plane of the paper. Emf induced across the ends of the rod is e=BAn =BÏ€l2n = (Bl2Ï€)/T = 1/2 Bl2Ï‰

Q8. Fan is based on
•  Electric Motor
•  Electric dynamo
•  Both
•  None of these
Solution
It is a fact.

Q9.The number of turns of primary and secondary coils of a transformer are 5 and 10 respectively and the mutual inductance of the transformer is 25 henry. Now the number of turns in the primary and secondary of the transformer are made 10 and 5 respectively. The mutual inductance of the transformer in henry will be
•  6.25
•  12.5
•  25
•  50
Solution
M= (Î¼0N1N2A)/l

Q10.  The primary winding of a transformer has 100 turns and its secondary winding has 200 turns. The primary is connected to an ac supply of 120 V and the current flowing in it is 10 A. The voltage and the current in the secondary are
•  240 V,5 A
•  240 V,10 A
•  60 V,20 A
• 120 V,20 A
Solution

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