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Standard XII

Physics

Magnetic Field Due to a Straight Long Wire

Wires 1 and 2...

Question

Wires 1 and 2 carrying currents $i_{1}$ and $i_{2}$ respectively are inclined at an angle $θ$ to each other. What is the force on a small element dl of wire 2 at a distance of r from wire 1 (as shown in figure) due to the magnetic field of wire1

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Solution

The correct option is C
Length of the component dl which is parallel to wire(1) is $d l c o s θ$ , so force on it.
$F = \frac{μ_{0}}{4 π} . \frac{2 i_{1} i_{2}}{r} (d l c o s θ) = \frac{μ_{0} i_{1} i_{2} d l c o s θ}{2 π r}$

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Wires 1 and 2 carrying currents i1 and i2 respectively are inclined at an angle θ to each other. What is the force on a small element dl of wire 2 at a distance of r from wire 1 (as shown in figure) due to the magnetic field of wire1

The correct option is C Length of the component dl which is parallel to wire(1) is dl cosθ, so force on it. F=μ04π.2i1i2r(dl cosθ)=μ0i1i2dl cosθ2πr

Wires 1 and 2 carrying currents $i_{1}$ and $i_{2}$ respectively are inclined at an angle $θ$ to each other. What is the force on a small element dl of wire 2 at a distance of r from wire 1 (as shown in figure) due to the magnetic field of wire1

The correct option is C
Length of the component dl which is parallel to wire(1) is $d l c o s θ$ , so force on it.
$F = \frac{μ_{0}}{4 π} . \frac{2 i_{1} i_{2}}{r} (d l c o s θ) = \frac{μ_{0} i_{1} i_{2} d l c o s θ}{2 π r}$