Question

Two long parallel wires carrying currents 2.5 ampere 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 metre and 2 metre respectively from a collinear point R (see figure).

(i) An electron moving with a velocity of 4 x 10⁵ m/s along the positive x-direction experiences a force of magnitude 3.2 x 10^–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.

Solution

Given, two long parallel wires.
Current carried by first wire = 2.5 A
Current carried by second wire = 2.5 A
Distance of point P from R = 5 m
Distance of point Q from R = 2 m

(i) Magnetic field B₁ at R due to P is,

B_{1} = \frac{μ_{0} I}{2 π r} = \frac{μ_{0}}{2 π} (\frac{2.5}{5})

Magnetic field

B_{2}

at R due to Q is,

B_{2} = \frac{μ_{0}}{2 πr} \frac{I}{r} = \frac{μ_{0}}{2 π} (\frac{I}{2})

Therefore,
Resultant magnetic field at R is given by,

B = B_{1} + B_{2} = \frac{μ_{0}}{2 π} (\frac{2.5}{5} + \frac{I}{2}) . . . . . . (1)

This Resultant field at R acts downwards and perpendicular to PX.

Now,
Force experienced by electron moving along PX is

\vec{F} = e (\vec{v} \times \vec{B}) [θ = 90 °]

\vec{F}

is perpendicular to both

\vec{v} and \vec{B} .

Because of the negative charge of electron, the force F acts perpendicular to the plane of paper directed upwards.

Now,

F = evB

∴

B = \frac{F}{ev} = \frac{3.2 \times 10^{- 20}}{1.6 \times 10^{- 19} \times 4 \times 10^{5}} = 0.5 \times 10^{- 6}

...(2)

Equating this to (1), we get

\frac{μ_{0}}{2 π} [\frac{2.5}{5} + \frac{I}{2}] = 0.5 \times 10^{- 6}

which gives

I = 2 [\frac{2 π}{μ_{0}} (0.5 \times 10^{- 6}) - \frac{2.5}{5}] = 2 [\frac{2 π \times (0.5 \times 10^{- 6})}{4 π \times 10^{- 7}} - \frac{2.5}{2}]

= 2 [\frac{2.5}{1} - \frac{2.5}{5}] = 4 A

(ii) In this case, we can consider the following alternatives:

(a) If 'r' is the distance of the current 2.5 A which, is directed into the plane of paper from point R then,
we have

B_{3} = (\frac{μ_{0}}{2 π}) (\frac{2.5}{r})

Now,

B_{1} + B_{2} + B_{3} = 0

Therefore,

\frac{μ_{0}}{2 π} (\frac{2.5}{5} + \frac{4}{2} + \frac{2.5}{r}) = 0

which gives, r= -1 m .
Thus the third wire is located at 1 m from R on RX.

(b) When the current 2.5 A is directed out from the plane of paper in an upward direction.

Therefore,
I₃ = – 2.5 A

Hence, we will have

\frac{μ_{0}}{2 π} (\frac{2.5}{5} + \frac{4}{2} + \frac{2.5}{r}) = 0

which gives r = 1 m i.e., the third wire is located 1 m from R on RQ.

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