(A) The magnetic field $B$ produced by the first wire at a distance $l$ is given by $B = \frac{\mu_0 i_1}{2 \pi l}$.Since the wires are perpendicular,

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(A) The magnetic field $B$ produced by the first wire at a distance $l$ is given by $B = \frac{\mu_0 i_1}{2 \pi l}$.Since the wires are perpendicular,the angle between the magnetic field and the current $i_2$ is $90^{\circ}$.The magnetic force $dF$ on a small length $d$ of the second wire is given by $dF = i_2 (B) d \sin(90^{\circ})$.Substituting the value of $B$,we get $dF = i_2 \left( \frac{\mu_0 i_1}{2 \pi l} \right) d (1) = \frac{\mu_0 i_1 i_2 d}{2 \pi l}$.Therefore,the magnetic force is proportional to the product of the currents,$i_1 i_2$.

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

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Two long wires with no contact are placed perpendicular to each other. $i_1$ and $i_2$ are currents flowing through these wires respectively. The magnetic force on a small length '$d$' of the second wire situated at a distance '$l$' from the first wire is proportional to,

TS EAMCET 2021 All TS EAMCET

A
$i_1 i_2$
B
$l$
C
$\frac{1}{i_1 i_2}$
D
$l^2$

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Moving Charges and Magnetism

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Force on a Current Carrying Conductor

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Two long wires with no contact are placed perpendicular to each other. $i_1$ and $i_2$ are currents flowing through these wires respectively. The magnetic force on a small length '$d$' of the second wire situated at a distance '$l$' from the first wire is proportional to,

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$A$ and $B$ are two infinitely long straight parallel conductors. $C$ is another straight conductor of length $1 \, m$ kept parallel to $A$ and $B$ as shown in the figure. Then the force experienced by $C$ is

$A$ loop of flexible conducting wire lies in a magnetic field of $2.0 \,T$ with its plane perpendicular to the field. The length of the wire is $1 \,m$. When a current of $1.1 \,A$ is passed through the loop, it opens into a circle. The tension developed in the wire is (in $\,N$)

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