CBSE Class 12 · 2011 · All India · Magnetic Effects of CurrentShort Answer

A steady current

I_{1}

flows through a long straight wire. Another wire carrying a steady current

I_{2}

in the same direction is kept close and parallel to the first wire. Show with the help of a diagram how the magnetic field due to the current

I_{1}

exerts a magnetic force on the second wire. Write the expression for this force.

Answer: The two parallel wires attract; the force per unit length is $\frac{F}{ℓ} = \frac{μ _{0} I _{1} I _{2}}{2 π r}$ .

Solution

Wire 1 sets up a field

B_{1} = \frac{μ _{0} I _{1}}{2 π r}

at the location of wire 2, directed perpendicular to wire 2. Wire 2, carrying

I_{2}

, then feels a force

F = B_{1} I_{2} ℓ = \frac{μ _{0} I _{1} I _{2}}{2 π r} ℓ

, so

\frac{F}{ℓ} = \frac{μ _{0} I _{1} I _{2}}{2 π r} .

By Fleming's left-hand rule the force is attractive because the currents are in the same direction.

CBSE Class 12 Board · 2011 · All India · Subjective