Faraday's Law MCQ Level - 2 - Physics MCQ

Let I  be the current be in the outer coil 2.
The field at centre 
The flux through the inner coil due to current in the outer coil 2,

So, Mutual inductance of outer coil

Flux across the outer coil,

Let us consider there are n  turns of a coil having radius a and current  I. Then magnetic field at the centre of the coil is given by.

Flux of magnetic field through a coil.

Flux through n  coil

Self inductance given by relation

If number of turns in the loop is increased from 1 to 8 then, Self inductance increased by 64 time.
The correct answer is: 64L

Flux through small coil is

Hence, emf  is lay behind the large coil current by π/2
The correct answer is: lags the current in the large coil by π/2

In the presence of uniform magnetic field B perpendicular to the disc in outward direction (say) Force on a unit positive charge

  will lie in radially outward direction)
Work done in moving a unit positive charge from centre to the rim is equal to potential difference ε

The correct answer is:  

Let wire cd  is at distance x  from the edge be considered a plane surface bounded by the loop acdb  and unit normal vector    lying in same direction as 

Flux through area  acdb

= Blv
Since, Flux is increasing with time, emf will be induced so as to circulate current resulting in weakening of magnetic field. So, current should flow from P  to  Q.
Current

Lorentz force acting on rod 
= ilB  are perpendicular
 (forward up the incline)

Free body diagram of slide is shown in figure.

There is no acceleration in direction perpendicular to plane of the loop.

So, N = O   [N  Normal reaction perpendicular to the plane]

But in the plane of the loop, the wire cd  slide down due to gravitational force acting downward and lorentz force acting upward.

So, In downward direction

But when we talk about steady speed, 

Let at any instant of time, velocity of the rod is v  towards right. The current in the circuit is i. In the figure,

V_a – V_b = V_d – V_c

i.e. Ldi = Bldx
Li = Blx

Magnetic force on the rod at this instant is

Since, this force is in opposite direction of  so from Newton’s second law we an write

Comparing this with equation of SHM, i.e.,

Let us calculate motional emf produced in the rod. Consider a unit positive charge on the rod at a distance r  from the centre. Let rod rotates with angular velocity ω. Velocity of this unit positive charge in magnetic field in transverse direction is ωr.

Lorentz force acting on the unit charge

Potential difference between A  and  O = V_AO

work done in moving a unit positive charge from O  to A.

The correct answer is:  

To find out the flux through the loop. Consider an element of length a and breadth dx at a distance x  from wire.

∴  Area = adx = dS

Let the orientation of the loop is such that    lie in same direction,    is unit normal to surface.

Magnetic field due to long current carrying conductor wire.

at a distance x is given by

Flux of magnetic field through  element

When the flux changes in the second coil due to change in the current in first coil, then the mutual inductance is given by the relation.

Similarly, when flux changes in first coil due to change in the current in second coil.

From the Reciprocity Theorem
M₂₁ = M₁₂ = M

From (iii) and (iv)

Multiply (v) and (vi)
 

Let the distance of slider from AB is y. Consider an element of length y and width dz  at a distance z from the y axis.

dS = ydz

consider a length dz of the rod

∴  Magnetic force on rod dF = iB·dz