Question

Two long wires carrying same currents are placed
perpendicular to each other. The magnetic force on
a small length dL of second wire is​

Answer 1

Answer:

The magnetic force on small length of second wire is

$F = i(dL)(\frac{\mu_0 i}{2\pi r})$

Explanation:

magnetic field due to first wire at some distance "r" from it is given as

$B = \frac{\mu_0 i}{2\pi r}$

here we know that

i = current in the wire

$\mu_0 = 4\pi \times 10^{-7}$

now the force on second wire due to this magnetic field is given as

$F = i(L\times B)$

$F = i(dL)B sin 90$

$F = i(dL)(\frac{\mu_0 i}{2\pi r})$

so above is the expression for the force on small element of the wire

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Topic : Force on current carrying wire

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