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Two long parallel wires carrying current 2.5 amperes and I ampere I 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 4 x 10 5 m/s along the positive x - direction experience 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.

Two long parallel wires carrying current 2.5 amperes and I ampere I 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 4 x 105 m/s along the positive x  - direction experience 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. 

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Grade:upto college level

2 Answers

Navjyot Kalra
askIITians Faculty 654 Points
9 years ago
Hello Student,
Please find the answer to your question
. (i) The magnetic field (du to current in wire P ) at R
= μ0 / 4π x 2Ip / rPR = μ0 / 4π x 2 x 2.5 / 5
= μ0 / 4π [in the plane of paper downwards]
Similarly, the magnetic field (due to current is wire Q) at R
= μ0 /4π x 2 x I / 2 = μ0 /4π I
[in the plane of paper downwards]
The total magnetic field at R [due to P and Q]
B = μ0 / 4π + μ0 /4πI = μ0 / 4π (1 + I)
[in the plane of paper downwards]
The force experienced by the electron
F = qvB sin θ
= ev B sin 90° = 1.6 x 10-19 x 4 x 105 x μ0 / 4π (1 + I)
But F = 3.2 x 10-20 N (Given )
∴ 3.2 x 10-20 = 1.6 x 10-19 x 4 x 105 x 10-7 (1 + I )
⇒ I = 4 amp.
(ii) Let us consider a position between Q and R. The magnetic field produced should be equal 5 x 10-7 T in the plane of paper acting upwards.
For this let the wire having current 2.5 amp be placed at a distance r from R and current flowing outwards the plane of paper.
∴ 5 x 10-7 = μ0 / 4π x 2 x 2.5 / or r = 1 m
Let us consider another position beyond R collinear with P, Q and R. Let it be placed at a distance r’ from R, having current in the plane of paper.
∴ 5 x 10-7 = μ0 / 4π x 2 x 2.5 / or r’ = 1 m
Thanks
Navjot Kalra
askIITians Faculty
Chandrasekar SM
13 Points
4 years ago
(ii)There are two possibilities .The current carrying maybe either inward or outward.
a)The current is inward .Consider a point at a distance r from point R towards the direction of X.
Then i3=+2.5 A.It is given that Bnet at R is 0
Bnet=B1+B2+B3
The magnetic induction is same for all three wires 
B=(u0/2Π)(i/r)
Bnet=(u0/2Π)(i1/r1+i2/r2+i3/r)
Bnet=0
-i3/r=2.5/5+1/2
r=-2.5i.e.,2.5 m away from R in the negative direction of RX.
b)Current is outward.The wire now is placed at a distance r fom R on RQ.We get,
i3/r=i1/r1+i2/r2
We will get r=2.5 away from R on RQ .
 
 
 
 
 
 
 

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