NEET Physics: Topic-wise Test- 5 Free Online Test 2026

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.

V_DC=I_DCR
∴R=V_DC/I_DC=12/4=3Ω
I_AC=V_AC/Z=V_AC√(R²+X_L²)
2.4=12√ [(3)²+X_L²]
solving this equation, we get,
X_L=4Ω
X_C=1/ω_C=1/(50×2500×10⁻⁶)
=8Ω
Z=√[R2+(XC−XL)2] =5Ω
∴I=V_DC/Z=12/5=2.4A=I_rms
P=I_rms²R=(2.4)²(3)
=17.28W
At given frequency X_C>X_L if omega is further decreased XC will increase (as X_C∝1/ω) and X_L will increase (as X_L∝ω).
Therefore X_C−X_L and hence Z will increase. So, current will decrease.

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.
 

B= 2μ₀​i⋅πr²/4π(r²+x²)^3/2
Magnetic field at centre= μ₀​i/2r
B= μ₀​i/2r
Put r= √3
B¹=2μ_0​i⋅πr²/4π(4r²)^3/2
=B/8
Hence, √3R distance from the centre magnetic field is equal to magnetic field at centre 
 

The magnetic field at the center O due to the upper side of the semicircular current loop is equal and opposite to that due to the lower side of the loop.

No force is exerted by a magnetic field on a stationary electric dipole because:

• The electric dipole consists of two equal and opposite charges.
• When stationary, the dipole does not experience any motion through the magnetic field.
• As a result, there is no induced current or change in magnetic flux.
• The magnetic field only affects moving charges or current-carrying conductors.

Therefore, the correct understanding is that a magnetic field does not exert a force on a stationary electric dipole.

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

According to NCERT, the force on a current-carrying conductor in a magnetic field is given by

The direction is found using the right-hand rule for vector (cross) product: point your fingers along l (direction of current) and curl them towards B; your thumb gives the direction of force.

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

As the two identical coaxial circular loops carry a current i each circulating in the same direction, their flux add each other. When the two are brought closer, the total flux linkage increases. By Lenz's law a current will be induced in both the loops such that it opposes this change in flux. As the flux increases, the current in each loop will decrease so as to decrease the increasing flux.

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 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

Magnetic flux, φ = 12t² + 7t + 3 (in milliweber),
where t is in seconds.

The induced emf (E) is given by Faraday's law as the negative rate of change of magnetic flux:

E = - dφ/dt

Differentiate φ with respect to t:

dφ/dt = d/dt (12t² + 7t + 3) = 24t + 7

At t = 5 s,

dφ/dt = 24 × 5 + 7 = 120 + 7 = 127 mV

Therefore, the induced emf is

E = - dφ/dt = -127 mV

Hence, the induced emf at t = 5 s is -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.

The voltage can be written as,
V=5[sinωt+(π/2​)]
The angle between the voltage and the current is 90∘.
The power dissipated in the circuit is given as,
P=V_rms​I_rms​cosϕ
=V_rms​I_rms​cos90o
=0
Thus, the power dissipated in the circuit is 0.
 

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

V_R​=√ [V₂−V_C²​​]=√ [(10)²−(8)²]​=6V
tanϕ= X_C/X_R​ ​​= V_C​​/V_R​ =8/6​=4/3​

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