Physics: Topic-wise Test- 5 - NEET MCQ

By Faraday's Law of induction,
ε=− dϕ​/dt
=−Bl (dx/dt) ​=−Blv_0​
This emf should induce the movement of charges creating a current. But due to the attached capacitor, all charges are conserved.
Thus I= dq/dt ​=0
The correct option is C.

The magnetic field near the conducting rod due to long current carrying wire is 
B= μ₀​I/2πr ​ (using Ampere's law, ∫B.dl=μ_0​I)
Now, induced emf due to moving rod is e=Blv= (μ₀​I/2πr) ​×lv= μ₀​vIl/2πr

cosθ=sin(90^o−θ)
sinα+sinβ=2sin (α+β/2)​cos (α−β​/2)
i=2sin100πt+2cos(100πt+30^o)
=2sin100πt+2sin(90^o−(100πt+30^o))
=2sin100πt+2sin(60^o−100πt)
=2 x 2 x sin [{(100−100)πt+60^o }/2] x cox[(100+100)πt−60^o/2]
​=4 x sin30o∗cos(100πt−30^o)
=4x(1/2) x cos(100πt−30^o)
=2cos(100πt−30^o)
therefore, I_o​=2A
I_rms​=2 / √2​= √2​A

From Biot-Savart law, magnetic field at a point p, B= (μ0​/4π)∫ [(Idl×r​)/ r3]
where r is the distance of point p from conductor and I is the current in the conductor.
Thus magnetic field due to current carrying conductor depends on the current flowing through conductor and distance from the conductor and length of the conductor.
 

The magnetic field due to a circular loop is given by:
B= μ₀​​2πi​/4πr
=10⁻⁷×2π×2/0.0157 ​
=8×10⁻⁵ Wb/m²

I₁=3A
I₂=8A
l=5m
r=0.1m
F=?
Force per unit length on the small conductor is given by:
f= μ_o2I₁I₂/4πr
Total force on the length of small conductor is
F=fl
F= μ_o2I₁I₂ l/4πr
F=4πx10^-7x2x3x8x5/4π x 0.1
F=2.4x10^-4N

The direction of force (motion) of a current carrying conductor in a magnetic field is given by Fleming’s left hand rule. 
It states that ‘ If we hold the thumb, fore finger and middle finger of the left hand perpendicular to each other such that the fore finger points in the direction of magnetic field, the middle finger points in the direction of current, then the thumb shows the direction of force (motion) of the conductor.
 

In an iron cored coil the iron core is removed so that the coil becomes an air cored coil. The inductance of the coil will decrease.

E=dx/dt =d(a(Tt-t²))/dt=a(T-2t)
I=E/R=a(T-2t)/R
Heat liberated,
△k=I²Rdt


=(a² /R)[T²t+(4t³/3)-2Tt²]^T₀
=△H=a²T³/3R

In SI units, permeability is measured in Henries per meter H/m or Hm−1.
Henry has the dimensions of [ML²T⁻²A⁻²].
Dimensions for magnetic permeability will be [ML²T⁻²A⁻²]/[L]=[MLT⁻²A⁻²]

Induced emf= d(flux)/dt​=R[dq​/dt]
⇒Δq= Δ(flux)​/R=2BA/R​

As magnetic field lines due to current carrying wire is along the surface and hence,
ϕ=B.A=BAcos90^O=0

The direction of the induced current is as shown in the figure, according to Lenz’s law which states that the indeed current flows always in such a direction as to oppose the change which is giving rise to it.

The direction of the current will be anticlockwise.
 
According to Lenz's law, the current in the coil will be induced in the direction that'll oppose the external magnetic field .
As the flux due to the external magnet is increasing ( as the N-pole is brought close ) , the coil will have to induce current in the anticlockwise direction to oppose this increase in flux and not in clockwise direction as that'll end up supporting the external flux.

The direction of these circular magnetic lines is dependent upon the direction of current. The density of the induced magnetic field is directly proportional to the magnitude of the current. Direction of the circular magnetic field lines can be given by Maxwell's right hand grip rule or Right handed cork screw rule.

The rod is moving towards the right in a field directed into the page.
Now, if we apply Fleming's right hand rule, then the direction of induced current will be from end X to end Y.
But, according to Lenz's law the emf induced in the rod will be such that it opposes the motion of the rod.
Hence, the actual emf induced will be from end Y to end X. So, the current will also flow from end Y to end X.
Now, using the convention of current end Y should be positive and end X should be negative.
So, correct answer is option b

Given :
            ϕ=12t²+7t+C
where, C is a constant, t is in sec, ϕ in milli-weber
The induced e.m.f. in the loop at t=5, on differentiating the equation with respect to 't'.
          dtdϕ​=dtd​(12t²+7t+C)
                =24t+7
 at            t=5
                dtdϕ​=24×(5)+7
                          =120+7
                          =127 mV
 

Initailly the loop taken have no current or no magnetic field associated with it when the magnet moved towards the loop the galvanometer show deflection showing the presence of current and current moving in loop also has magnetic field around it
 

The correct answer is B.To calculate the induced e.m.f when a wire is moved between the gap of two poles of a magnet, we can use Faraday's Law of electromagnetic induction.

Calculation:

- Given magnetic flux, Φ = 6 x 10^-3 Wb
- Time taken, t = 1 s

Formula:

The induced e.m.f can be calculated using the formula:

E = -ΔΦ/Δt

Substitute the given values:

E = - (6 x 10^-3 Wb) / (1 s)
E = - 6 x 10^-3 V
E = 6 mV

Therefore, the value of induced e.m.f when a wire is moved between the gap of two poles of a magnet in 1s is 6 mV.

When the two loops are brought together, the flux linking the two loops increases. By Lenz's law a current will be induced such that it opposes this change of flux . Therefore to oppose the increase in flux, the current in the two loops will decrease, so the flux decreases.

Flux of the magnetic field through the loop
ϕ=B.Acosωt
Emf= −dϕ/dt​=B.Aωsinωt
ωt=0 to π,  sinωt=+ve
ωt=π to2 π,  sinωt=−ve
Hence, the direction of induced e.m.f. changes once in 1/2​ revolution.
 

Given,
Inductance and capacitance are joined in series, and inductance L=0.5H,
And capacitance C=8μF=8×10⁻⁶F.
We know that,
ω=1/√(LC)
ω=1/√(8×10⁻⁶×0.5)=1/(2×10⁻³)=0.5×10³HZ=500HZ

Time taken by alternating current to reach maximum from zero is 4T​.
where T is the time period of the AC function.
T=1/f​, Frequency (f)=50 Hz.
So time taken to reach maximum=T/4​=1/4f​=1/(4×50)​=5×10⁻³ sec
I_RMS​=​I₀/√2​, I0​=Peak value of current.
Peak value of current I₀​=√2​×I_RMS​=√2​×10=14.14 A
 

Here, V_0​=283V,R=3Ω,L=25×10⁻³H
C=400μF=4×10⁻⁴F
Maximum power is dissipated at resonance, for which 
 
ν=1/[2π√(LC)]​​= (1×7​)/[2×22(25×10⁻³×4×10⁻⁴)]
=(7×10^3​)/(44√10​)
=50.3Hz

Here,
X_L = 4 Ω
R = 3 Ω
Z = √(X_L² + R²)=√(4² + 3²)
  = 5 Ω
E₀ = 10 V
In the LR circuit current in the ckt is given by
I = (E₀/Z)sin(ωt - Φ)
Φ = tan^-1(X_L/R)
=>  Φ = tan^-1(4/3)
I = (E_0/Z)sin(ωt - Φ)
  = (10/5) sin(ωt - Φ)
I at t = T/2
  = 2sin(ωT/2 – Φ)
 =  2sin(π - Φ)
  = 2sin(tan^-1(4/3))
  = 2×0.8
  = 1.6 A
Potential difference across the resistor = 1.6×R
   = 1.6×3
   = 4.8 V
Potential difference across the inductor = 1.6XL
      = 1.6×4 = 6.4 V

Impedance= V_ac/I_ac​ ​​=100V/0.5A ​=200Ω
R= V_dc​​/I_dc ​ =100/1​=100Ω
Lω =√ (impedance²−R²​)=√(200²−100²​)=173.2
L=173.2/ ω​= 173.2/2π50​=0.55H
 

Peak value of voltage
V₀=i₀X_C =i₀/2πvC
=>1.57/(2x3.14x50x100x10^-6)
Hence if equation of current i = i_o sin ω t then in capacitive circuit voltage is 
V=V₀ (ωt -  π/2)
=>50[sin2π x50t- (π/2)]
 =50[100πt-(π/2)]

Given: V=V₀cosωt

Here,

∴I=I₀cos(ωt−π/6)

cosϕ₁=0.6=5/3
tanϕ₁=4/3= X_C/R₂…(1)
cosϕ₁=0.5=1/2
tanϕ₂=√3= R₂/X_C…(2)
From (1) and (2)3√3/4=R₁/R₂