Q 1 :

The figure shows certain wire segments joined together to form a coplanar loop. The loop is placed in a perpendicular magnetic field in the direction going into the plane of the figure. The magnitude of the field increases with time. I1 and I2 are the currents in the segments ab and cd. Then,          [2009]

  • I1>I2

     

  • I1<I2

     

  • I1 is in the direction ba and I2 is in the direction cd

     

  • I1 is in the direction ab and I2 is in the direction dc

     

(4)

The magnetic field is increasing in the downward direction. Therefore, according to Lenz's law the current I1 will flow in the direction ab and I2 in the direction dc.



Q 2 :

A small bar magnet is being slowly inserted with constant velocity inside a solenoid as shown in the figure. Which graph  best represents the relationship between emf induced with time?                  [2004]

  •  

  •  

  •  

  •  

(3)

Polarity of emf will be opposite when the magnet enters and leaves the coil.

Only graph (3) shows these characteristics.



Q 3 :

A rectangular conducting loop of length 4 cm and width 2 cm is in the xy-plane, as shown in the figure. It is being moved away from a thin and long conducting wire along the direction 32x^+12y^ with a constant speed v. The wire is carrying a steady current I=10 A in the positive x-direction. A current of 10μA flows through the loop when it is at a distance d=4 cm from the wire. If the resistance of the loop is 0.1Ω, then the value of v is ______ ms-1.

[Given: The permeability of free space μ0=4π×10-7 N A-2]                  [2023]



(4)

 Here e.m.f., ε=(B1-B2)bvy

Current, i=εR=μ0I2πR(1d-1d+a)bvy

10-5=2×10-7×100.1[14-18]×2·vy

[Given: I=10 A, R=0.1Ω, i=10μA=10-5 A]

 vy=2

tanθ=vyvx13=2vx

 vx=23

 v=vx2+vy2=(23)2+(2)2

        =16=4 ms-1



Q 4 :

A series R-C combination is connected to an AC voltage of angular frequency ω=500 rad/s. If the impedance of the R-C circuit is R1.25, the time constant (in millisecond) of the circuit is                         [2011]



(4)

Time constant, T=RC

Impedance Z=R2+(1ωC)2

Given Z=R1.25

  R1.25=R2+(1ωC)2

 RC=2ω=2500×1000 ms

 RC=4 ms



Q 5 :

A small circular loop of area A and resistance R is fixed on a horizontal xy-plane with the center of the loop always on the axis n^ of a long solenoid. The solenoid has m turns per unit length and carries current I counterclockwise as shown in the figure. The magnetic field due to the solenoid is in n^ direction. List-I gives time dependences of n^ in terms of a constant angular frequency ω. List-II gives the torques experienced by the circular loop at time t=π6ω. Let α=A2μ02m2I2ω2R.             [2022]

  List-I   List-II
(I) 12(sinωtj^+cosωtk^) (P) 0
(II) 12(sinωti^+cosωtj^) (Q) -α4i^
(III) 12(sinωti^+cosωtk^) (R) 3α4i^
(IV) 12(cosωtj^+sinωtk^) (S) α4j^
    (T) -3α4i^

Which one of the following options is correct?

  • (I) → (Q); (II) → (P); (III) → (S); (IV) → (T)

     

  • (I) → (S); (II) → (T); (III) → (Q); (IV) → (P)

     

  • (I) → (Q); (II) → (P); (III) → (S); (IV) → (R)

     

  • (I) → (T); (II) → (Q); (III) → (P); (IV) → (R)

     

Select one or more options

(3)

(I)   B=μ0mI2(sinωtj^+cosωtk^)

ϕ=B·A=μ0mI2cos(ωt)A

ε=-dϕdt=μ0mIωA2sin(ωt)

i=εR=μ0mIωA2Rsin(ωt)

So, M=iA=μ0mIωA22Rsin(ωt)k^

τ=M×B=μ02m2I2ωA22Rsin2(ωt)(-i^)

For t=π6ω

τ=μ02m2I2ωA22R×14(-i^)

=A2μ02m2I2ω2R×14(-i^)=α4(-i^).  So, (I)Q

(II) Solving as (I), ϕ=0, ε=0, i=0, M=0. So, (II)P

Similarly, (III)S

                  (IV)R



Q 6 :

A current carrying infinitely long wire is kept along the diameter of a circular wire loop, without touching it, the correct statement(s) is(are)         [2012]

  • The emf induced in the loop is zero if the current is constant.

     

  • The emf induced in the loop is finite if the current is constant.

     

  • The emf induced in the loop is zero if the current decreases at a steady rate.

     

  • The emf induced in the loop is infinite if the current decreases at a steady rate.

     

Select one or more options

(1, 3)

If the current is constant, the emf induced in the loop is zero. Emf will be induced in the circular wire loop when flux through it changes with time.

e=-ΔϕΔt

When the current is constant, the flux changing through it will be zero.

Also, if the current decreases at a steady rate, the emf induced in the loop is zero. When the current is decreasing at a steady rate then the change in the flux (decreasing inwards) on the right half of the wire is equal to the change in flux (decreasing outwards) on the left half of the wire such that Δϕ through the circular loop is zero.