First slide

Force on a current carrying wire placed in a magnetic field

Question

Two infinitely current carrying insulated wires AB and CD are carrying same current I. The wires cross each other at right angles as shown in figure. Three identical conductors P₁Q₁ , P₂Q₂ and P₃Q₃, carrying same current are placed a shown in figure and the forces acting on them are F₁ , F₂and F₃ respectively. Then select the correct option.

Difficult

A

$F_{1} \neq 0$ and direction of $\vec{F_{1}}$ is wrong
B

$F_{2} \neq 0$ and direction of $\vec{F_{2}}$ is correct
C

$F_{3} \neq 0$ and direction of $\vec{F_{3}}$ is right
D

$F_{2} = F_{3} = 0$

Solution

Magnetic induction at any point on the conductor P, Q = $\frac{μ_{o} I}{2 πr} - \frac{μ_{o} I}{2 πr} = 0$ .
There fore F₁ = 0.
Magnetic induction at any point on the conductor P₂Q₂ is given by $B_{2} = \frac{μ_{o} I}{2 πr} + \frac{μ_{o} I}{2 πr} = \frac{μ_{o} I}{πr}$ and its direction is outward. Now $\vec{F_{2}} = i \vec{P_{2} Q_{2}} \times \vec{B_{2}} .$
Hence the direction of $\vec{F_{2}}$ shown in the figure is right. Again magnetic induction at any point on the conductor P₃Q₃ is $B_{3} = \frac{μ_{o} I}{2 πr} + \frac{μ_{o} I}{2 πr} = \frac{μ_{o} I}{πr}$ and its direction is inward.
Since $\vec{F_{3}} = i \vec{P_{3} Q_{3}} \times \vec{B_{3}}$ , direction of $\vec{F_{3}}$ shown in the diagram is wrong.

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