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201.

An electron and a proton have equal kinetic energies. They enter in a magnetic field perpendicularly, ThenA. Both will follow a circular path with same radiusB. Both will follow a helical pathC. Both will follow a parabolic pathD. All the statements are false

Answer» Correct Answer - D
Particles entering perpenducylary, hence they will describe circular path. Since their masses are different so they will describe path of different radii.
Discussion
202.

A homogenous electric field `E` and a uniform magnetic field `vecB` are pointing in the same direction. A proton is projected with its velocity parallel to `vecE`. It willA. Go on moving in the same direction with increasing velocityB. Go on moving in the same direction with constant velocityC. Turn to its rightD. Turn to its left

Answer» Correct Answer - A
Here magnetic force is zero, but the velocity increases due to electric force.
Discussion
203.

A U tube of uniform square cross-sectional side a has mercury in it. Current I is passed between sealed electrodes x and y. A magnetic field `B` is applied across horizontal section perpendicular to the plane of diagram. The difference in mercury levels is [Given: `rho` = density of mercury] .A. `2/3 (BI)/(rho g a)`B. `(2BI)/(rho g a)`C. `(BI)/(rho g a)`D. `(4BI)/(rho g a)`

Answer» Correct Answer - c
`(hrhog)a^2=Bia or h=(BI)/(rhoga)`
Discussion
204.

A `U`-shaped wire of mass `m` and length `l` is immersed with its two ends in mercury (see figure). The wire is in a homogenous field of magnetic induction `B`. if a charge, that is, a current pulse `q=int idt`, set through the wire, the wire will jump up. If the wire reaches height `h` calculate the magnitude of the charge or current pulse, assuming that the time of the current pulse is very small in comparision with the time of flight. Make use of the fact that impulse of force equals `intFdt`, which equal `mv`. Evalute `q` for `B=0.1 Wb//m^(2), m=10 gm,l=20cm & h=3` meters.A. `sqrt(24)C`B. `sqrt(48)C`C. `sqrt(15)C`D. None of these

Answer» Correct Answer - C
Mercury is a conductor so current can flow through mercury.
`F=ilB`
`intFdt=mv`
`intilBdt=msqrt(2gh)`
`q=intidt=(msqrt(2gh))/(lB)`
`=(10xx10^(-3)xxsqrt(2xx10xx3))/(20xx10^(-2)xx0.1)=sqrt(15)C`
Discussion
205.

A square loop of wire carrying current I is lying in the plane of paper as shown in Fig. 1.141. The magnetic field is present in the region as shown. The loop will tend to rotate A. about PQ with KL coming out of the pageB. about PQ with KL going into the pageC. about RS with MK coming out of the pageD. about RS with MK going into the page

Answer» Correct Answer - a
See the direction of torque about centre.
Discussion
206.

A long straight wire of radius `a` carries a steady current `i`. The current is uniformly distributed across its cross section. The ratio of the magnetis field at `(a)//(2) and (2a)` isA. `1//4`B. `4`C. `1`D. `1//2`

Answer» Correct Answer - C
At `r=a/2, B_(1)=(mu_(0)Ir)/(2pia^(2))=(mu_(0)I(a//2))/(2pia^(2))=(mu_(0)I)/(4pia)`
`r=2a, B_(2)=(mu_(0)I)/(2pir)=(mu_(0)I)/(4pia)`
`(B_(1))/(B_(2))=1`
Discussion
207.

Two long parallel copper wires carry currents of `5 A` each in opposite directions. If the wires are separated by a distance of ` 0.5m` , then the force between the two wires isA. `10^(-5)N`, attractiveB. `10^(-5)N` repulsiveC. `2xx10^(-5)N`, attractiveD. `2xx10^(-5)N`,repulsive

Answer» Correct Answer - B
`F=10^(-7)(2i_(1)i_(2))/(a)=10^(-7)xx(2xx5xx5)/(0.5)=10^(-5)N` (Repulsive)
Discussion
208.

A particle of mass `M` and charge `Q` moving with velocity `vec(v)` describe a circular path of radius `R` when subjected to a uniform transverse magnetic field of induction `B`. The work done by the field when the particle completes one full circle isA. `BQv2pi R`B. `((Mv^(2))/(R)) 2pi R`C. zeroD. `BQ 2 pi R`

Answer» Correct Answer - C
`W = F.d cos 90^(@) = 0`
Discussion