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Magnetics question

2023 · 24 Jan · Shift 1 · Q51

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2023 · 24 Jan · Shift 1 · Q51

Magnetics question

2023 · 24 Jan · Shift 1 · Q51

JEE MainPhysicsMagneticsMCQ+4 / −1

Two long straight wires P and Q carrying equal current 10A each were kept parallel to each other at 5 cm distance. Magnitude of magnetic force experienced by 10 cm length of wire P is F

_1

. If distance between wires is halved and currents on them are doubled, force F

_2

on 10 cm length of wire P will be:

A
$\frac{F_1}{8}$
B
10 F $_1$
C
$\frac{F_1}{10}$
D
8 F $_1$

View written solutionFree

Correct answer: D

Magnetic force between two parallel current-carrying wires

For two long parallel wires separated by distance $d$ , carrying currents $I_1$ and $I_2$ , the force on length $L$ of one wire is

F = \frac{\mu_0 I_1 I_2 L}{2\pi d}

So,

F \propto \frac{I_1 I_2}{d}

Initial force $F_1$

Initially,

$I_1 = I_2 = 10\,\text{A}$
distance $d = 5\,\text{cm}$
length $L = 10\,\text{cm}$

Thus,

F_1 \propto \frac{(10)(10)}{5}

New conditions for $F_2$

Now,

distance is halved: $d' = \frac{d}{2}$
currents are doubled: $I_1' = 2I_1$ , $I_2' = 2I_2$

Then,

F_2 \propto \frac{(2I_1)(2I_2)}{d/2}

F_2 \propto \frac{4 I_1 I_2}{d/2}

F_2 \propto 4 I_1 I_2 \cdot \frac{2}{d} = 8\frac{I_1 I_2}{d}

Therefore,

F_2 = 8F_1

Option check

A: $\frac{F_1}{8}$ ❌
B: $10F_1$ ❌
C: $\frac{F_1}{10}$ ❌
D: $8F_1$ ✅

So the correct option is D.

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