# 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.

^{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.

## 2 Answers

_{0}/ 4π x 2I

_{p}/ r

_{PR}= μ

_{0}/ 4π x 2 x 2.5 / 5

_{0}/ 4π [in the plane of paper downwards]

_{0}/4π x 2 x I / 2 = μ

_{0}/4π I

_{0}/ 4π + μ

_{0}/4πI = μ

_{0}/ 4π (1 + I)

^{-19}x 4 x 10

^{5}x μ

_{0}/ 4π (1 + I)

^{-20}N (Given )

^{-20}= 1.6 x 10

^{-19}x 4 x 10

^{5}x 10

^{-7 }(1 + I )

^{-7}T in the plane of paper acting upwards.

^{-7}= μ

_{0}/ 4π x 2 x 2.5 / or r = 1 m

^{-7}= μ

_{0}/ 4π x 2 x 2.5 / or r’ = 1 m

Navjot Kalra

askIITians Faculty

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

_{net}=B

_{1}+B

_{2}+B

_{3}

_{0}/2Π)(i/r)

_{net}=(u

_{0}/2Π)(i

_{1}/r

_{1}+i

_{2}/r

_{2}+i

_{3}/r)

_{net=0}

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

_{3}/r=i

_{1}/r

_{1}+i

_{2/}r

_{2}