Test: Motional EMF - JEE MCQ

θ=20 cos πt + t² + 50t + 25m Wb
We know that,
Emf = dθ/dt = -20 x π sinπt+2t+50+0
At,t = 1
Emf = -20 πsin π+2+50
       = 0 + 2 + 50
       = 52 mV

At the poles of the copper rod, the earth’s magnetic field will be more. This is explained with the help of Lorentz force where he explains that the magnetic field is generated due to the movement of charges.
 
Earth’s magnetic field is defined as the magnetic dipole which is similar to bar magnet and is tilted 11° from the earth’s axis of rotation.
 

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'.
        We know that,  E = - dϕ / dt

So,  d​(12t²+7t+C) / dt
                =  24t+7
 at            t = 5
                 E ​= 24×(5)+7
                          = 120+7
                          = 127 mV

So, E = - dϕ / dt = - 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.
 

ε = 100 sin 100πV
ε = ε₀ sinωt
ω=100π
2π n= 100π
n = 50rps

Motional emf= Bℓv
It depends on magnetic field intensity, length and velocity. The diameter doesn't affect any of the factors.

As the copper has high electrical conductivity. Hence, higher e.m.f would be induced in the copper loop.