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.
$B=B_P+B_Q$
$=\frac{{\mu }_0}{2\pi }\frac{I_P}{5}+\frac{{\mu }_0}{2\pi }\frac{I_Q}{2}=\frac{{\mu }_0}{2\pi }(\frac{2.5}{5}+\frac{I}{2})$
$=\frac{{\mu }_0}{4\pi }(I+1)={10}^{-7}(I+1)$
Net force on the electron will be,
$F_m=Bqvsin90^{\circ }$
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.