A rectangular wire frame of dimensions `(0.25 xx 2.0 m)` and mass 0.5 kg falls from a height 5m above a region occupied by uniform magnetic field of magnetic induction 1 T. The resistance of the wire frame is `1/8 (Omega)`. Find

time taken by the wire frame when it just starts coming out of the magnetic field.

Options

A. 0.2s
B. 1s
C. 2.2s
D. `sqrt((1)/(5))s`

Answer

C

Explanation

(a) Velocity of wire frame when it starts entering the magnetic field.
`V_(1) sqrt(2 gh) = sqrt(2(10)(5))= 10m//s`
and the time taken is `t, = sqrt((2h)/(g) = sqrt((2(5))/(10))=1s` .
(ii) When the frame has partially entered the field, the induced emf produced is `epsilon =Blv`
`I=(epsilon)/(R) = (Blv)/(R)` (anticlockwise)
Ampere's force, `F=(B^(2)l^(2)v)/(R)` (upward)
Putting `m=0.5 kg, B=1 T, l=0.25m,
`v=10m//s`
R=1//8 Omega`
We get, `F=((1)^(2)(0.25)^(2)(10))/(1//8) =5N`
Since, `mg=(0.5)(10) = 5N`

Since, `mg=(0.5)(10)=5N`
Therefore, using Newton's second law, the acceleration of the wire frame while entering the magnetic field is zero, thus, time taken to completely entering into the field is `t_(2)=2/10 =0.2 s`
(iii) When the frame has completely entered the field, current become zero and thus, the ampere's force also become zero. The frame accelerates under gravity only.
`:. 15 = 10t_(3)+5t_(3)2`
or, `t_(3)^(2)+2t_(3)-3 =0` or `t_(3)=1s`
The total time taken is
`T=t_(1)+t_(2)+t_(3)=1 + 0.2 + 1 = 2.2 s`.