### Magnetic Effects Of Current Question 1

#### Question 1 - 24 January - Shift 1

Two long straight wires $P$ and $Q$ carrying equal current $10 A$ each were kept parallel to each other at $5 cm$ distance. Magnitude of magnetic force experienced by $10 cm$ length of wire $P$ is $F_1$. If distance between wires is halved and currents on them are doubled, force $F_2$ on $10 cm$ length of wire $P$ will be :

(1) $8 F_1$

(2) $10 F_1$

(3) $F_1 / 8$

(4) $F_1 / 10$

## Show Answer

#### Answer: (1)

#### Solution:

#### Formula: Force Per Unit Length Between Two Parallel Wires Carrying Currents

Force per unit length between two parallel straight

wires $=\frac{\mu_0 i_1 i_2}{2 \pi d}$

$\frac{F_1}{F_2}=\frac{\frac{\mu_0(10)^{2}}{2 \pi(5 cm)}}{\frac{\mu_0(20)^{2}}{2 \pi(\frac{5 cm}{2})}}=\frac{1}{8}$

$\Rightarrow F_2=8 F_1$