Question 91 of 276

Three straight parallel current carrying conductors are shown in the figure. The force experienced by the middle conductor of length $25 \mathrm{~cm}$ is:

A$3 \times 10^{-4} \mathrm{~N}$ toward rightB$6 \times 10^{-4} \mathrm{~N}$ toward rightC$9 \times 10^{-4} \mathrm{~N}$ toward rightDZero

Show answer and solution▾

Answer: (A) $3 \times 10^{-4} \mathrm{~N}$ toward right

$$\text { } I_1=30 \mathrm{~A} \quad I=10 \mathrm{~A} \quad I_2=20 \mathrm{~A}$$


Also given; length of wire $Q$ $=25 \mathrm{~cm}=0.25 \mathrm{~m}$
Force on wire $\mathrm{Q}$ due to wire $\mathrm{R}$
$$\begin{aligned} &F_{Q R}=10^{-7} \times \frac{2 \times 20 \times 10}{0.05} \times 0.25 \\ &=20 \times 10^{-5} \mathrm{~N} \text { (Towards left) } \end{aligned}$$
Force on wire $\mathrm{Q}$ due to wire $\mathrm{P}$
$$\begin{aligned} &F_{Q P}=10^{-7} \times \frac{2 \times 30 \times 10}{0.03} \times 0.25 \\ &=50 \times 10^{-5} \mathrm{~N} \text { (Towards right) } \\ &\text { Hence, } F_{\text {net }}=F_{Q P}-F_{Q R} \\ &=50 \times 10^{-5} \mathrm{~N}-20 \times 10^{-5} \mathrm{~N} \\ &=3 \times 10^{-4} \mathrm{~N} \text { towards right } \end{aligned}$$

JEE Main Physics practice question on Magnetic Effects & Magnetism (Magnetic Field), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.