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1.

In a cyclotron, a charged particleA. undergoes acceleration all the time.B. Speeds up between the dees because of the magnetic fieldC. speeds up in a dae.D. slows down within a dee and speeds up between dees.

Answer» Correct Answer - A
The charged particle undergoes acceleration as
i. speeds up betwee the dees because of the oscillating electric field and
(ii). Speed remain the same inside the dees because of the magnetic field bu direction undergoes change continuously.
Discussion
2.

A long straight wire carrying current of `25A` rests on a table as shown in figure. Another wire PQ of length 1m, mass `2*5g` carries the same current but in the opposite direction. The wire PQ is free to slide up and down. To what height will PQ rise?

Answer» The magnetic field produced by long straight wire carrying current of 25A rests on a table on small wire `B=(mu_(0)I)/(2pih)`
the magnetic force on small conductor is
`F=Bilsintheta=Bil`
Force applied on PQ balance the weight of small current carrying wire.
`F=mg=(mu_(0)I^(2)l)/(2pih)`
`h=(mu_(0)I^(2)l)/(2pimg)=(4pixx10^(-7)xx25xx25xx1)/(2pixx2.5xx10^(-3)xx9.8)=51xx10^(-4)`
`h=0.51cm`
Discussion
3.

Two long wires carrying current `I_(1) and I_(2)` are arranged as shown in figure the one carrying current `I_(1)` is along is the x-axis. The other carrying current `I_(2)` is along a line parallel to the y-axis given by x=0 and z=d. Find the force exerted at `O_(2)` because of the wire along the x-axis.

Answer» In Biot-Savart law, magnetic field B is parallel to idl`xx`r and idl have its direction along the direction of flow of current.
Here. For the direction of magnetic field, At `O_(2)` due to wire carrying `I_(1)` carrying is
`B||"parallel "idlxxr" or "hatixxhatk" but "hatixxhatk=-hatj`
So, the direction at `O_(2)` is along Y-direction
The direction of magnetic force exerted at `O_(2)` because of the wire along the x-axis
`F=IlxxB=hatjxx(-hatj)=0`
So, the magnetic field due to `l_(1)` is along the y-axis the second wire is along the y-axis and hence, the force is zero.
Discussion
4.

A multirange current meter can be constructed by using a galvanometer circuit as shown in figure. We want a current meter that can measure `10mA`, `100mA` and `1A` using a galvanometer of resistance `10Omega` and that produces maximum deflection for current of `1mA`. Find `S_1, S_2` and `S_3` that have to be used.

Answer» `I_(G).G=(I_(1)-I_(G))(S_(1)+S_(2)+S_(3))` for `I_(1)=10mA`
`I_(G)(G+S_(1))=(I_(2)-I_(G))(S_(2)+S_(3))` for `I_(2)=100mA`
and `I_(G)(G+S_(1)+S_(2))=(I_(3)-I_(G))(S_(3))` for `I_(3)=1A`
gives `S_(1)=1W,S_(2)=0.1W`
and `S_(3)=0.01W`
Discussion