Two long parallel wires carrying current ` 2.5 amperes` and `I ampere` in the same direction ( directed into the plane of the paper) are held at `P and Q` respectively such that they are perpendicular to the plane of paper. The points `P and Q` are located at a distance of ` 5 metres and 2 metres` respectively from a collinear point `R` ( see figure)
(i) An electron moving with a velocity of ` 4xx10^(5) m//s ` along the positive ` x- direction ` experiences a force of magnitude `3.2xx10^(-20) N` at the point `R`. Find the value of `I`.
(ii) Find all the positions at which a third long parallel wire carrying a current of magnitude `2.5 amperes` may be placed so that the magnetic induction at `R` is zero.

Explanation

(a) Magnetic field at R due to both wires P and Q will be
downward as shown in Fig. Therefore, net field at
R will be sum of these two.
`B=B_P+B_Q=(mu_0I)/(2pi) (I_P)/5+(mu_0I)/(2pi) (I_Q)/2=(mu_0)/(2pi)(2.5/5+I/2)`
`=(mu_0)/(4pi)(I+1)=10^-7(I+1)`

Net force on the electron `F=Bqv sin 90^@`
or `(3.2xx10^-20)=(10^-7)(I+1)(1.6xx10^-19)(4xx10^5)`
or `I+1=5 :. I=4A`
(b) Net field at R due to wires P and Q is
`B=10^-7(I+1)T=5xx10^-7T`
Magnetic field due to third wire carrying a current of 2.5A
should be `5xx10^-7T` in upward direction so that net field
at R becomes zero.
Let the distance of this wire from R be r.
Then,
`(mu_0)/(2pi) 2.5/r=5xx10^-7 or ((2xx10^-7)(2.5))/r=5xx10^-7m`
or `r=1m`
So, the third wire can be put at
M or N as shown in Fig.
If it is placed at M, then cur-
rent in it should be outwards
and if placed at N, then cur-
rent will be inward.