Search any question & find its solution
Question:
Answered & Verified by Expert
A long wire lies along $X$-axis and carries a current of $40 \mathrm{~A}$ in positive $x$-direction. A second long wire is perpendicular to the $x y$-plane, passes through point $(3.0 \mathrm{~m}) \hat{j}$ and carries a current along positive $z$-direction. If the magnitude of resultant magnetic field at the point $(2.0 \mathrm{~m}) \hat{j}$ $R=5 \times 10^{-6} \mathrm{~T}$ then the current in the second wire is
(Permeability of free space, $\mu_0=4 \pi \times 10^{-7}$ SI unit)
Options:
(Permeability of free space, $\mu_0=4 \pi \times 10^{-7}$ SI unit)
Solution:
1076 Upvotes
Verified Answer
The correct answer is:
$15 \mathrm{~A}$
Given arrangement is
There are two magnetic fields at point $2 \hat{j}$ as shown
$B_1=$ magnetic field due to wire $A$.
$=\frac{\mu_0 I_1}{2 \pi d}=\frac{\mu_0 \times 40}{2 \pi \times 2} \mathrm{~T}$
and $B_2=$ magnetic field due to wire $B$
$=\frac{\mu_0 I_2}{2 \pi d}=\frac{\mu_0 \times I_2}{2 \pi \times 1} \mathrm{~T}$
Resultant field magnitude,
$\begin{aligned} R & =\sqrt{B_1^2+B_2^2} \\ & =\frac{\mu_0}{2 \pi} \sqrt{20^2+I_2} \quad \text { (Given, } R=5 \times 10^{-6} \mathrm{~T} \text { ) }\end{aligned}$
$\begin{aligned} & \Rightarrow \sqrt{20^2+I_2^2}=\frac{5 \times 10^{-6}}{2 \times 10^{-7}} \\ & \Rightarrow 400+I_2^2=625\end{aligned}$
$I_2^2=225$
$\Rightarrow \quad I_2=15 \mathrm{~A}$
There are two magnetic fields at point $2 \hat{j}$ as shown
$B_1=$ magnetic field due to wire $A$.
$=\frac{\mu_0 I_1}{2 \pi d}=\frac{\mu_0 \times 40}{2 \pi \times 2} \mathrm{~T}$
and $B_2=$ magnetic field due to wire $B$
$=\frac{\mu_0 I_2}{2 \pi d}=\frac{\mu_0 \times I_2}{2 \pi \times 1} \mathrm{~T}$
Resultant field magnitude,
$\begin{aligned} R & =\sqrt{B_1^2+B_2^2} \\ & =\frac{\mu_0}{2 \pi} \sqrt{20^2+I_2} \quad \text { (Given, } R=5 \times 10^{-6} \mathrm{~T} \text { ) }\end{aligned}$
$\begin{aligned} & \Rightarrow \sqrt{20^2+I_2^2}=\frac{5 \times 10^{-6}}{2 \times 10^{-7}} \\ & \Rightarrow 400+I_2^2=625\end{aligned}$
$I_2^2=225$
$\Rightarrow \quad I_2=15 \mathrm{~A}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.