JEE Main · 2019 · 10 Apr 2019 · Shift 1 · Magnetic Effects of Current

Two wires

A

and

B

are carrying currents

I_{1}

and

I_{2}

as shown in the figure. The separation between them is d . A third wire C carrying a current I is to be kept parallel to them at a distance x from A such that the net force acting on it is zero. The possible values of

x

are:

(A) $x = (\frac{I _{1}}{I _{1} - I _{2}}) d$ and $x = \frac{I _{2}}{( I _{1} + I _{2} )} d$
(B) $x = \pm \frac{I _{2} d}{( I _{1} - I _{2} )}$ ✓ Correct
(C) $x = (\frac{I _{2}}{I _{1} + I _{2}}) d$ and $x = (\frac{I _{2}}{I _{1} - I _{2}}) d$
(D) $x = (\frac{I _{1}}{I _{1} + I _{2}}) d$ and $x = (\frac{I _{2}}{I _{1} - I _{2}}) d$

Answer: (B)

Solution

Place the third wire C between A and B at a distance

x

from A, leaving

d - x

to B. The currents in A and B pull on C in opposite senses, and C feels zero net force when the two magnetic forces balance:

\frac{μ _{0} I _{1} I}{2 π x} = \frac{μ _{0} I _{2} I}{2 π ( d - x )}

. Cross-multiplying,

I_{1} (d - x) = I_{2} x

, which rearranges to

x = \frac{I _{1} d}{I _{1} + I _{2}}

. The null point lies between the wires, dividing the gap in the ratio

I_{1} : I_{2}

measured from A.

JEE Main · 2019 · 10 Apr 2019 · Shift 1 · Single correct option

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JEE Main · 2019 · 10 Apr 2019 · Shift 1 · Magnetic Effects of Current

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