(A) The force per unit length between two parallel current-carrying wires is given by $F = \frac{\mu_0 i_1 i_2}{2\pi r}$.For a wire of length $l$,the

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(A) The force per unit length between two parallel current-carrying wires is given by $F = \frac{\mu_0 i_1 i_2}{2\pi r}$.For a wire of length $l$,the total force is $F = \frac{\mu_0 i_1 i_2 l}{2\pi r} = \frac{\mu_0}{4\pi} \cdot \frac{2 i_1 i_2 l}{r}$.Given: $i_1 = 10 \, A$,$i_2 = 2 \, A$,$r = 10 \, cm = 0.1 \, m$,$l = 2 \, m$,and $\frac{\mu_0}{4\pi} = 10^{-7} \, T \cdot m/A$.Substituting the values:$F = 10^{-7} \times \frac{2 \times 10 \times 2 \times 2}{0.1} = 10^{-7} \times \frac{80}{0.1} = 10^{-7} \times 800 = 8 \times 10^{-5} \, N$.

Class 12 Physics Moving Charges and Magnetism Force on a Current Carrying Conductor

Medium

Through two parallel wires $A$ and $B$,$10 \, A$ and $2 \, A$ of currents are passed respectively in opposite directions. If the wire $A$ is infinitely long and the length of the wire $B$ is $2 \, m$,the force on the conductor $B$,which is situated at a $10 \, cm$ distance from $A$,will be:

A
$8 \times 10^{-5} \, N$
B
$4 \times 10^{-7} \, N$
C
$4 \times 10^{-5} \, N$
D
$4\pi \times 10^{-7} \, N$

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Moving Charges and Magnetism

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Force on a Current Carrying Conductor

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EasyAP EAMCET 2020

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MediumAIIMS 2019

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EasyKCET 2012

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DifficultAP EAMCET 2016

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Through two parallel wires $A$ and $B$,$10 \, A$ and $2 \, A$ of currents are passed respectively in opposite directions. If the wire $A$ is infinitely long and the length of the wire $B$ is $2 \, m$,the force on the conductor $B$,which is situated at a $10 \, cm$ distance from $A$,will be:

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The figure shows an equilateral triangle $ABC$ of side length '$l$' carrying currents as shown,placed in a uniform magnetic field '$B$' perpendicular to the plane of the triangle. The magnitude of the net magnetic force on the triangle is:

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