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QUESTION: 1

The current in inner coil is ** i = 2t^{2}**, as shown in figure. The current induced in the outer coil has a resistance of

Solution:

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,

QUESTION: 2

A circular loop of radius * a* is made of a single turn of thin conducting wire. The self inductance of this loop is

Solution:

Let us consider there are ** n** turns of a coil having radius a and current

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

QUESTION: 3

A large circular coil of ** N** turns and radius

Solution:

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

QUESTION: 4

A metal disc of radius * a* rotates with an angular velocity ω about an axis perpendicular to its plane passing through its centre in a magnetic field of induction

Solution:

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:

QUESTION: 5

A wire coil of length ** l**, mass

Solution:

Let wire ** cd** is at distance

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

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** [

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,

QUESTION: 6

A loop is formed by two parallel conductor connected by a solenoid with inductance ** L** and a conducting rod of mass

Solution:

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

*V _{a}* –

i.e.

Li

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.,

QUESTION: 7

A conducting rod of length ** a** is rotating with constant angular speed ω about its perpendicular bisector. A uniform magnetic field

Solution:

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

work done in moving a unit positive charge from ** O** to

The correct answer is:

QUESTION: 8

An infinite long wire carrying a current I(t) = I_{0} cos(ωt) is placed at a distance * a* from a square loop of side

Solution:

To find out the flux through the loop. Consider an element of length * a* and breadth

∴ 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

QUESTION: 9

Two coil of self inductance ** L_{1}** and

Solution:

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

*M*_{21} = *M*_{12} = *M
*

From (iii) and (iv)

Multiply (v) and (vi)

QUESTION: 10

Two parallel rails separated by a distance l are connected through a resistance ** R** in

Solution:

Let the distance of slider from ** AB** is

*dS* = *ydz*

consider a length ** dz** of the rod

∴ Magnetic force on rod *dF* = *iB*·*dz*

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