**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**

A) 1V

B) 11V

C)9V

D)10V

(C)

e _{induced } = – dΦ / dt = – d/dt(BA) = -A dB/dt

= (l^{2} / 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**

(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**

(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**

4. ( b)

Explanation:

B_{1 }= (μ_{0} i / 2 ) ( 2R)^{2 } / 2 [ (2R)^{2 }+ (2R)^{2 }]^{3/2} ; B_{2 }= μ_{0/2 } iR^{2 }/ 2 [ R^{2 } + R^{2} ] ^{3/2}

B_{1 }_{/ 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.**

(D) The induced emf = E = dΦ /dt

= d/dt × (B.A) = A dB/dt

= (πr^{2} ) B_{0 } (d/dt )× ( e^{-t}) = – πr^{2} B_{0} e^{-t }

⇒ E_{0} = B_{0 }πr^{2} e ^{-t }| _{t=0 =B0 πr2 }

∴ The electrical power developed in the resister just at the instant of closing the key = p = E_{0}^{2 }/R =B_{0}^{2}π^{2} r^{4 }/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**

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^{2 }

∴ i= **Bωa**^{2}

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) 5B_{0} X^{2} Wb

B)3B_{0} X^{2} Wb

C)2B_{0} X^{2} Wb

D)B_{0} X^{2} Wb

8. (A)

Here,

→ →

A = x^{2 } κ^{ˆ} m^{2} and B = B_{0 }( 3i^{ˆ} +4j^{ˆ +5kˆ)} T

→ →

as Φ = B .A =B_{0 ( 3iˆ +4jˆ +5kˆ) .x2 kˆ}

**9. The phase relationship between current and voltage in a pure resistive circuit is best represented by**

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?**

10. (C)

Capacitive reactance, X_{c }= 1/ ωc = 1/ 2πνC ⇒ X_{c } = 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 :-**

12. (B)

B^{2} V_{0} L^{2} = mg ⇒ V = mgr

______ ____

R B^{2 }L^{2}

**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

(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

(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**

16. (a)

Explanation:

E × 2πr = πr ^{2} .α

∴ E = r 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 = V

X_{L } 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**

(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)

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

at t=0 , i_{1} (t) .i_{2}_{ } =0

also i_{1} (t) .i_{2 }(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 :**

(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?**

. (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} / 2R1 (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 I 4π R1Magnetic flux linked with the coil of radius R1 is**

**Φ = B1 π R2 ^{2 } N2 = μ0 /4π ( 2πN1 I /R1 ) π R2 ^{2 } N2 **

**Φ = MI = ∴ M = Φ /I = μ0 /4π ( 2πN1 /R1 ) π R2 ^{2 } N2 = μ0 π R2 ^{2 }N1N2 / 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 ^{2 } B) ^{3} / _{2 Bωl }^{2}**

**C) _{ Bωl }^{2 }D) 2 _{Bωl }^{2}**

** **

**24. (A)Consider a small element of the rod of length dx at a distancex 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 isde = Bvdx = BωxdxThe emf across the entire length lis given by **

** e=∫de = Bω _{0}∫^{l} xdx = Bωl^{2} / 2 **

**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= πr ^{2 }B**

**Now induced emf e is given by e = – dΦ /dt = -d/dt(πr ^{2 }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)17Compare V =150sin (150 t) with V= V0 sin ωt, we get V0 =150V**

**Compare I =150 sin ( 150t+ π/3 ) with I= I 0 sin (ωt+Φ ) , we get I = 150 A, Φ = π/3 =60ºThe 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 HzTHEREFOR 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.**