Physics 01 – Electro Magnetic Induction

1. A square loop of side 1 m is placed in a perpendicular magnetic field. Half of the area of the loop lies inside the magnetic field. A battery of emf 10 V and negligible internal resistance is connected in the loop. The magnetic field changes with time according to the relation B = (0.01 – 2t) tesla. The resultant emf of the circuit is

AC

A) 1V 

B) 11V

C)9V

D)10V

(C) 

induced  = – dΦ / dt = – d/dt(BA) = -A dB/dt 

= (l2 / 2  )    ( d(0.01 -2t ) / dt  )   = 1 volt 

∴Resultant emf =10 -1 =9v

2. The direction of induced current in the right loop in the situation shown by the given figure is

AC 1

(A) along the common axis                                         (B) along xzy
(C) along xyz                                                                    (D) none of these 

2. (C) The induced current in the right loop will be along xyz.

3. The variation of induced emf (ε) with time (t) in a coil if a short bar magnet is moved along its axis with a constant velocity is best represented as

ASAAS

(B) The polarity of emf will be opposite in the two cases while the magnet enters the coil and while the magnetic leaves the coil. Only in option (B) polarity is changing.

4. Figure shows two coils carrying equal currents. The ratio of  magnetic field at P due to coil-1 to coil-2 is

AC 2

4. ( b)

Explanation: 

B1   = (μ0  i  / 2 )              ( 2R) /  2  [ (2R)2  + (2R)2  ]3/2     ;        B2   =         μ0/2      iR2  / 2       [ R2 + R2 ] 3/2

 

              B1    /   B 2     =  (4/8 )3/2  /   (1/2 √2  )=    4× 2√2 / 8×2√2         =1:2

                                               

5.Two circular coils can be arranged in any of three situations as shown in the figure. Their mutual inductance will be

(A)maximum in situation (i)                                              (B) maximum in situation (ii)

(C) maximum in situation (iii)                                           (D)same in all situations

(A)  A mutual inductance occurs when the magnetic field generated by a coil induces a voltage in a secondary coil.

M= ε m    / (  dI/dt  )

Since, in option (A), there is maximum emf is induced in secondary coil. So, mutual inductance is maximum.

6. Shown in the figure is a circular loop of radius r and resistance R. A variable magnetic field of induction B = B is established inside the coil. If the key (K) is closed, the electrical power developed right after closing the switch is equal.

AC 3

(D)   The induced emf = E = dΦ /dt

                         = d/dt  × (B.A) = A dB/dt

                          = (πr2 )  B (d/dt )× ( e-t) = – πr2 B0 e-t  

⇒ E0 = Bπr2 e -t   | t=0     =Bπr2  

∴ The electrical power developed in the resister just at the instant of closing the key  = p = E02  /R   =B02π2 r4  /R

7. The arrangement shown here is a Faraday’s disc of radius a which is rotating in a uniform magnetic field B perpendicular to the plane of disc. The current passing through the resistance is
AC 4

7. (b)
Explanation: Emf induced in the disc will be same as that induced in a rod rotating in a magnetic field.  

∴ e= 1/2  ×  Bωa  

∴ i=  Bωa2

          2R

8. A square of side x m lies in the x-y plane in a region, where the magnetic field is given bywhere B0 is constant. The magnitude of flux passing through the square

A) 5B0 X2 Wb

B)3B0 X2 Wb

C)2B0 X2 Wb

D)B0 X2 Wb

8. (A)

Here,

→                       →             

A = x κˆ m2   and B = B0  ( 3iˆ +4jˆ +5kˆ)  T 

            → →              

as Φ = B .A =B( 3iˆ +4jˆ +5kˆ) .x

9. The phase relationship between current and voltage in a pure resistive circuit is best represented by
AC 6
ASAAS 2

9. (C)
In the pure resistive circuit current and voltage both are in phase. Hence graph (C) is correct.

10. Which of the following graphs represents the correct variation of capacitive reactance XC with frequency?10. Which of the following graphs represents the correct variation of capacitive reactance XC with frequency?
AC 7

10. (C)
Capacitive reactance, X=  1/ ωc  =  1/ 2πνC  ⇒ X  = 1/ν With increase in frequency, XC decreases.
Hence, option (C) represents the correct graph.

11. The natural frequency (ωο) of oscillations in LC circuit is given by

11. (C)

12. A rod PQ mass ‘m’ and length can slide without friction on two vertical conducting semi-infinite rails. It is given a velocity Vο downwards, so that it continues to movedownward  6 with the same speed Vο on its own at any later instant of time. Assuming g to be constant every where, the value of Vο is :-
AC 8

12. (B) 

 

 B2 V0 L2  = mg ⇒ V = mgr

______                             ____   

   R                                 BL2

13. Alternating voltage (V) is represented by the equation  

(A)V (t) = Vm e ωt                 (B) V (t)= Vm sin ωt

 (C) V (t)= Vm cot ωt                (D) V (t)= Vm tan ωt

where Vm  is the peak voltage

13. (B)
The equation of the alternating voltage is
V( t)  = v m   sin ωt

14. In the case of an inductor

(A)voltage lags the current by  π/2
(B) voltage leads the current by π/2
(C) voltage leads the current by π/3
(D)voltage leads the current byπ/4

14. (B)
In an inductor voltage leads the current by π/2   or current lags the voltage by π/2 .

15. The magnetic flux linked with a coil of N turns of area of cross section A held with its plane parallel to the field B is

(A) NAB /2                                                      (B) NAB
(C)NAB  /4                                                      (D) zero

15. (D)
Magnetic flux linked with a coil
Φ= NBA cosθ
Since the magnetic field B is parallel to the area A, i.e.,θ= 90°.
∴Φ = 0

16. In the cylindrical region shown, magnetic field is diminishing at the rate of  α (T/s). The force on the electron at a distance r along y-axis is
AC 9

16. (a)

 

Explanation:  

    E × 2πr = πr 2  .α 

    ∴  E = α 

               2

17. An ideal inductor is in turn put across 220 V, 50 Hz and 220V, 100 Hz supplies. The current flowing through it in the two cases will be

(A) equal                                                              (B) different
(C) zero                                                                 (D)infinite </strong></span>

An ideal inductor is in turn put across 220 V, 50 Hz and 220 V, 100 Hz supplies. The current (B) The current in the inductor coil is given by   
 I =    =   V      
        XL       2πνL   
Since frequency ν in the two cases is different, hence the
current in two cases will be different.

18. An ac source of voltage V = Vm sin ωt is connected across the resistance R as shown in figure. The phase relation between current and voltage for this circuit is 
AC 10

(A)both arc in phase
(B) both are out of phase by 90°
(C) both are out of phase by 120°
(D)both are out of phase by 180°

18. (A) The given circuit is a pure resistive circuit. In this circuit the voltage and current both are in phase

19. A coil of wire having finite inductance and resistance has a conducting ring placed coaxially within it. The coil is connected to a battery at time t = 0, so that a time-dependent current I1(t) starts flowing through the coil. If I2(t) is the current induced in the ring, and B(t) is the magnetic field at the axis of coil due to I1(t), then as a function of time (t > 0), the product I2(t) B(t)

(A) increases with time
(B) decreases with time
(C) does not vary with time
(D) passes through a maximum.

(D) 

2    (t)   B(t)  ∝  i 2  (t) i (t) ∝ ( 1- e-t/τ ) e-t/τ  

  at t=0 , i1   (t) .i2    =0  

also i1   (t) .i2   (t) → 0 as t → ∞ 

20. In figure when key is pressed the ammeter A reads i ampere. The charge passing in the galvanometer circuit of total resistance R is Q. The mutual inductance of the two coils is :
AC 11

(A) Q/R                                                                        (B) QR
(C) QR/i                                                                       (D) i/QR 

(C) 

21. A small bar magnet is being slowly inserted with constant velocity inside a solenoid as shown in figure. Which graph best represents the relationship between emf induced time?
AC 14

. (C)

22. Two circular coils of radii R1 and R2 having N1 and N2 turns are placed concentrically in the same plane. If R2<

(A) μ0  πR2 2 / 2R                                                 (B) μ0  πR2 2 N1N2/ 2R1   

(C)    μ0  πR2  N1N2/ 2R1                                      (D)  μ0  πR1  N1N2/ 2R2  

. (B)
Let I be the current through the coil of radius R1. The magnetic induction at the centre of the coil is

B = μ0    2πN1
         4π       R1
Magnetic flux linked with the coil of radius R1 is

Φ = B1 π R2  N2    =   μ0 /4π  2πN1 I /R1 ) π R2  N2 

Φ = MI = ∴ M = Φ /I =  μ0 /4π  2πN1  /R1 ) π R2  N2    =  μπ RN1N2  / 2R1

23. The frequency for which a 5.0μF capacitor has a reactance of 1000Ω  is given by

(A)1000   cycles/ sec
        π                               

(B) 100 cycles/ sec
        π 
(C) 200 cycle /s

(D)5000 cycles /sec

(B) 

Xc  = 1/ ωc  

⇒    ω = 1/ Xc C =  1 / 1000×5×10-6  

f = 1/ 2π × 5 × 10-3   = (100 /π )cycle /sec

                                                 

24. A copper rod of length l  is rotated about one end perpendicular to the uniform magnetic field B with constant angular velocity ω. The induced emf between the end is

A)  ½  B ωl                                                                 B)  32  Bωl 2

C)  Bωl 2                                                                                             D) 2 Bωl 2

 

24. (A)
Consider a small element of the rod of length dx at a distance
x from the centre O. Let v be the linear velocity of the  element at the right angle to the magnetic field B. The emf developed across the element is

de = Bvdx = Bωxdx
The emf across the entire length l
is given by  

              e=∫de = Bω  0l  xdx = Bωl2  / 2                      

AC 13
25. A conducting circular loop is placed in a uniform magnetic field of induction B tesla with its plane normal to the field. Now the radius of the loop starts shrinking at the rate (dr/dt). Then, the induced emf at the instant when the radius is r, is

A) πrB (dr /dt ) 

B) 2πrB (dr /dt ) 

C)πr 2 (dB /dt ) 

D) πr 2 /2   B(dr /dt ) 

(B) 

if the radius is r at a time t then the ins flux  Φ is given by  Φ=  πr2  B

Now induced emf e is given by    e = – dΦ /dt = -d/dt(πr2  B )  = -πB ( 2r dr/dt)

                                                               =  -2 πBr dr/dt 

induced emf =2 πBr dr/dt  ( numerically)

 

INGTEGERS TYPE QUESTION

26. Find the mutual inductance if number of turns in primary and secondary coils is increased to two times each.

26. (4 times)
M =.μ0   N1 N2 A   /  l

∴ M becomes 4 times

27. In an ac circuit, V and I are given by V= 150 sin (150 t) V and  I= 150 sin ( 150t + π ⁄ 3 )A.Find the power dissipated in the circuit.

27. (5625 W)
17
Compare V =150sin (150 t)  with V= Vsin ωt, we get V0  =150V

Compare I =150 sin ( 150t+ π/3 ) with I= I sin (ωt+Φ ) , we get I = 150 A, Φ  = π/3 =60º

The power dissipated in ac circuit is   p =1/2 v0 I 0   Cos Φ = 1/2 × 150 1/2  × 1/2 = 5625w 

28. A 100 Ω resistor is connected to a 220 V, 50 Hz ac supply. What is the rms value of current in the circuit?

(2.8 A)
Here, R = 100 Ω, Vrms = 220 V, υ= 50 Hz
THEREFOR  Irms=  V ms/ R  = 220/100 =2.2A

29. Find the capacitive reactance if a 5 μF capacitor is connected to a 200 V, 100 Hz ac source. ?

(318 Ω)

30. A conductor is moving with the velocity  ν in the magnetic field and induced current is I. If the velocity of conductor becomes double, the induced current will be _________

(2I)
When the velocity of conductor becomes double, area intercepted becomes twice. Therefore induced current becomes twice.

ABOUT US

NERAM classroom is Tamil nadu’s one of the top NATA and JEE coaching center which made great change in students especially from small towns. Started in 2017,and successfully we have alumni who have been placed under top institutes such as NIT ,SPA ,AU and some other good institutes under Anna universities. NERAM classroom offers highly personalised and effective learning programs for aspirants of competitive exams like JEE, NATA etc . NERAM classroom has become one of the most preferred education platforms across India.

 

WHY WE ARE INCOMPATIBLE ?

Our niche is creating personalised learning experiences for every type of learner. NERAM breaks away from a typical classroom teaching structure and uses technology and customisation to enable students to interact with their tutors.In India, students often feel too shy to clarify their doubts in a crowded coaching class. Many students travel long hours to get to class. NERAM breaks these shackles by liberating learning and teaching from the limits of location.Students are given full freedom to interact with teachers,experts and alumni as per their convenient time.

 

WHY OPT US ?

Traditionally, students rely on brick-and-mortar coaching centers, but companies like NERAM classroom are springing up and more students are supplementing offline courses with online tutoring.
NERAM classroom is an incredibly personalized tutoring platform for you, while you are staying at your home. We have grown leaps and bounds to be the best Online Tuition Website in India with immensely talented Master Teachers, from the most reputed institutions. We provide you year-long structured coaching classes for JEE and NATA entrance exam preparation at affordable tuition fees, with an exclusive session for clearing doubts, ensuring that neither you nor the topics remain unattended. Revision notes and formula sheets are shared with you, for grasping the toughest concepts. Assignments, Regular Homeworks, Subjective & Objective Tests promote your regular practice of the topics.

 

PERSONALISED LEARNING AND TUTORING

Traditionally, students rely on brick-and-mortar coaching centers, but companies like NERAM classroom are springing up and more students are supplementing offline courses with online tutoring.
Personalised Learning is the foundation of our unique programs for students. Using knowledge graphs, the program adapts and creates personalised learning journeys for students.Based on the student's progress, personal learning profiles are created which help analyse strengths and areas of improvement. Each concept gets tagged at different levels of difficulty to create a smooth learning curve which keeps students motivated through any new learning challenge. Whenever errors are made, motivating remedial methods are generated to strengthen and improve the student's learning experience.

 

TECHNOLOGY ENABLED LEARNING

At NERAM, we leverage technology to merge best practices like use of well verified study materials, videos, engaging content and quizzes with the best teachers so that every student across India has access to the best learning and guidance for their future .

KEY FEATURES OF NERAM CLASSROOM

  • INTERACTIVE LEARNING SESSIONS
  • NUMBER OF MOCK TESTS
  • FRIENDLY GUIDE
 

BEST TEACHERS & ENGAGING CONTENT

NERAM classroom Learning Programs provide students a holistic learning experience. Students across regions can access the best teachers.Very few teams like BYJU’S, VEDANTA in India have a great sense for the education sector as well as how technology can be used for effective scaling, NERAM is one such team.
Hours and Hours of Study with no fun, is a bad idea for you, foreseeing the long run. To ensure that motivation is stirred in the best proportion for your clear understanding, a good number of quizzes and Objective tests are organized to impart knowledge. FREE Sample Papers and Important questions are extracted, solved and discussed, ensuring that you are 100% prepared before any exam.
Hours and Hours of Study with no fun, is a bad idea for you, foreseeing the long run. To ensure that motivation is stirred in the best proportion for your clear understanding, a good number of quizzes and Objective tests are organized to impart knowledge. FREE Sample Papers and Important questions are extracted, solved and discussed, ensuring that you are 100% prepared before any exam.

  Website Contents © NeramClasses. All rights reserved.