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1.

shown in the figure is a small loop that is kept co-axially with the bigger loop. If the slider moves from `A` to `B`, then A. Current flow in both the loops will be in opposite senses.B. Clockwise current in loop `1` and anti-clockwise current in loop `2` flowC. No current flows in loop `2`D. Clockwise current flows in loop `2`

Answer» When the slider moves towards `B`, the resistance of the circuit (bigger loop) decrease. Therefore, the current in the bigger loop increases. The increasing current results in increasing flux `(phi alpha i)` linked in the smaller coil. Consequently, induced emf will be generated along the smaller loop creating an induced current so as to oppose the increase in flux. Therefore the current flows anticlockwise, in the inner loop.
Discussion
2.

A conducting loop is pulled with a constant velocity towards a region of unifrom maganetic field of induction `B` as shown in the figure. Then the current involved in the loop is `(d gt r)` A. clockwise while enteringB. anti-clockwise while enteringC. zero when completely insideD. clockwise while leaving

Answer» When the loop is drawn into the magnetic filed, the area of the portion loop in the magnetic field will increase. That means, the flux linkage increases. Therefore a (vortex) emf is induced in the loop so as to oppose the change, that is to say to oppose the increase in magnetic flux in the loop. Therefore, the current will have to induce in anti-clockwise direction to induce an opposing magnetic field, that is pointing outward of the page. After the complete entry of the loop into the magnetic field, no variation of flux occurs. Therefore it induces no current so long as the loop . Is completely inside the magnetic field. When it emerges out of the magnetic field, following the previous argument, the direction of induced current in it will be reversed (clockwise)
Discussion
3.

The coefficient of mutual inductance of the two coils is `0.5 H`. If the current is increased from `2` to `3` `A ln 0.01 sec`. In one of them, then the induced e.m.f. in the second coil is :A. `25V`B. `50V`C. `75V`D. `100V`

Answer» Induced emf `=M(di)/(dt)=0.5(3-2)/(0.01)=50V`
Discussion
4.

A resistor `R`, an inductor `L` and a capacitor `C` are connected in series to a source of frequency `n`. If the resonant frequency is `n_(r)`, then the current lags behind voltage whenA. `n=0`B. `n lt n_(r)`C. `n=n_(r)`D. `n gt n_(r)`

Answer» Below resonant frequency the current leads the applied emf, at resonance it is in phase with applied emf and above resonance frequency it lags the applied emf.
Discussion
5.

An ideal choke takes a current of `8A` when connected to an a.c. source of `100` volt and `50Hz`. A pure resistor under the same conditions takes a current of `10A`. If two are connected in series to an `a.c.` supply of `100V` and `40Hz`, then the current in the series combination of above resistor and inductor isA. `10A`B. `8A`C. `5sqrt(2)`ampD. `10sqrt(2)`amp

Answer» `X_(L)=(100)/(8)`, `R=(100)/(10)=10Omega`
`Lxx100pi=(100)/(8)` or `L=(1)/(8pi)H`
`Z=sqrt(((1)/(8pi)xx2pixx40)^(2)+10^(2))=10sqrt(2)`
`I=(E)/(Z)=(100)/(10sqrt(2))=5sqrt(2)A`
Discussion
6.

A conducting loop of resistance `R` and radius `r` has its center at the origin of the co-ordinate system in a magnetic field of induction `B`. When it is rotated about `Y`-axis through `90^(@)`, net charge flown in the coil is directly proportional to A. `B`B. `R`C. `r^(2)`D. `r`

Answer» Induced emf loop when the variation of flux `d phi` during time `dt` is given as
`E=(d phi)/(dt)rArr int_(phi_(1))^(phi_(2))d phi=Delta phi=int Edt`……….`(1)`
`rArr` The total charge flown in the loop `=q=int I dt`
`q=int (E)/(R ) dt`………..`(2)`
using `(1)` and `(2)`
`q=(Delta phi)/( R )`
where `Delta phi=` change in flux given as `Delta phi=phi_(2)-phi_(1)=B.(pir^(2))` because intially to flux is linked with the coil and it has maximum flux linkage `phi_(2)=B pir^(2)` when turned through `90^(@)`, the flux is reduced to zero.
`rArr q=(piBr^(2))/(R )`
i.e. `q prop BrArr q prop r^(2) rArr q prop (1//R)`
Discussion