Two long parallel wires X and Y, separated by a distance of 6 cm, carry currents of 5 A and 4 A, respectively, in opposite directions as shown in the figure. Magnitude of the resultant magnetic field at point P at a distance of 4 cm from wire Y is $x \times 10^{- 5} T$ . The value of x is _____. Take permeability of free space as $μ_{0} = 4 π \times 10^{- 7}$ SI units. [2025]

Question

Smart Achievers · Accepted Answer

(1)

At P

$x \times 10^{- 5} T$

$\vec{B} = \frac{μ_{0}}{2 π} (\frac{4}{4} \times 10^{2} - \frac{5}{10} \times 10^{2}) \hat{k}$

$= 1 \times 10^{- 5} T$

$\Rightarrow x = 1$

Solution

Ans.
(1)

At P

$\vec{B} = {\vec{B}}_{1} + {\vec{B}}_{2} = \frac{μ_{0} i_{2}}{2 π r_{2}} \hat{k} - \frac{μ_{0} i_{1}}{2 π r_{1}} \hat{k}$

$\vec{B} = \frac{μ_{0}}{2 π} (\frac{4}{4} \times 10^{2} - \frac{5}{10} \times 10^{2}) \hat{k}$

$= 1 \times 10^{- 5} T$

$\Rightarrow x = 1$

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Solution

Ans. (1) At P B→=B→1+B→2=μ0i22πr2k^–μ0i12πr1k^ B→=μ02π(44×102–510×102)k^ =1×10–5 T ⇒ x=1

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Ans.
(1)

At P

$\vec{B} = {\vec{B}}_{1} + {\vec{B}}_{2} = \frac{μ_{0} i_{2}}{2 π r_{2}} \hat{k} - \frac{μ_{0} i_{1}}{2 π r_{1}} \hat{k}$

$\vec{B} = \frac{μ_{0}}{2 π} (\frac{4}{4} \times 10^{2} - \frac{5}{10} \times 10^{2}) \hat{k}$

$= 1 \times 10^{- 5} T$

$\Rightarrow x = 1$