First slide

Applications involving charged particles moving in a magnetic field

Question

Two long conductors, separated by a distance d carry current $I_{1}$ and $I_{2}$ in the same direction. They exert a force F on each other. Now the current in one of them is increased to two times and direction is reversed. The distance is also increased to 3d. The value of the force between them is

Moderate

A

-2F
B

F/3
C

-2F/3
D

–F/3

Solution

$C a s e (i) F = \frac{μ_{0} I_{1} I_{2}}{2 π d} (a t t r a c t i v e) C a s e (i i) F^{1} = \frac{μ_{0} (2 I_{1}) I_{2}}{2 π (3 d)} (r e p u l s i v e) \Rightarrow F^{1} = \frac{- 2}{3} F (n e g a t i v e \to r e p u l s i v e)$

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