Question

Two long parallel wires carrying currents 2.5 A 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 m and 2 m respectively from a collinear point R (see figure). a) An electron moving with a velocity of $ 4 \times 10^ 5 m/s$ along the positive x-direction experiences a force of magnitude $ 3.2 \times {10}^{-20} $N at the point R. Find the value of I. (b) Find all the positions at which a third long parallel wire carrying a current of magnitude 2.5 A may be placed, so that the magnetic induction at R is zero.

Answer 1

(a) Magnetic field at R due to both the wires P and Q will be
downwards as shown in figure.
Therefore, net field at R will be sum of these two.
$ \, \, \, \, \, \, \, \, \, \, \, \, \, B = B_P + B_Q$
$ \, \, \, \, \, \, \, = \frac{ {\mu}_0 }{2 \pi } \frac{ I_P}{ 5}+ \frac{ {\mu}_0 }{2 \pi } \frac{ I_Q}{ 2}= \frac{ {\mu}_0 }{2 \pi }\bigg( \frac{ 2.5}{5}+ \frac{I}{2} \bigg) $
$ \, \, \, \, \, \, \, \, \, \, \, \, \, = \frac{ {\mu}_0 }{4 \pi }(I+1) = {10}^{-7} (I+1) $
Net force on the electron will be,
$ \, \, \, \, \, \, F_m =Bqv sin 90 ^\circ $
or $ \, \, \, \, \, \, \, \, \, \, (3.2 \times {10}^{-20}) = (10^{-7})(I+1) $
$ \, \, \, \, \, \, \, \, \, \, \, \, (1.6 \times {10}^{-19})(4 \times 10^5 ) $
or $$ 7 + 1=5
$$ 7 = 4A
(b) Net field at R due to wires P and Q is
$$ $B = {10}^{-7} (I+1) T $
$ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, = 5 \times {10}^{-7} T$
Magnetic field due to third wire carrying a current of
2.5 A should be $ 5 \times {10}^{-7 }$ T in upward direction so, that
net field at R become s zero. Let distance of this wire from
R be r. Then,
$ \, \, \, \, \, \, \, \, \, \frac{{\mu}_0 }{2 \pi} \frac{ 2.5}{r} = 5 \times {10}^{-7}$
or $\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, =\frac{ (2 \times {10}^{-7}) (2.5)}{r} = 5 \times {10}^{-7} m $
or $\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, r= 1m$
So, the third wire can be put at M or N as shown in figure.
If it is placed at M, then current in it should be outwards
and if placed at N, then current be inwards.