# 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.

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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.

(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.

## 2 Answers

8 years ago

Hello Student,

Please find the answer to your question

. (i) The magnetic field (du to current in wire P ) at R

= μ

_{0}/ 4π x 2I_{p}/ r_{PR}= μ_{0}/ 4π x 2 x 2.5 / 5= μ

_{0}/ 4π [in the plane of paper downwards]Similarly, the magnetic field (due to current is wire Q) at R

= μ

_{0}/4π x 2 x I / 2 = μ_{0}/4π I[in the plane of paper downwards]

The total magnetic field at R [due to P and Q]

B = μ

_{0}/ 4π + μ_{0}/4πI = μ_{0}/ 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 10^{5}x μ_{0}/ 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 10^{5}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}= μ_{0}/ 4π x 2 x 2.5 / or r = 1 mLet 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}= μ_{0}/ 4π x 2 x 2.5 / or r’ = 1 mThanks

Navjot Kalra

askIITians Faculty

Navjot Kalra

askIITians Faculty

3 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 i

_{3}=+2.5 A.It is given that Bnet at R is 0B

_{net}=B_{1}+B_{2}+B_{3}The magnetic induction is same for all three wires

B=(u

_{0}/2Π)(i/r)B

_{net}=(u_{0}/2Π)(i_{1}/r_{1}+i_{2}/r_{2}+i_{3}/r)B

_{net=0}-i

_{3}/r=2.5/5+1/2r=-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,

i

_{3}/r=i_{1}/r_{1}+i_{2/}r_{2}We will get r=2.5 away from R on RQ .