Search any question & find its solution
Question:
Answered & Verified by Expert
A straight conductor of length $32 \mathrm{~cm}$ carries a current of $30 \mathrm{~A}$. Magnetic induction at a point in air at a perpendicular distance of $12 \mathrm{~cm}$ from the mid-point of the conductor is
Options:
Solution:
2696 Upvotes
Verified Answer
The correct answer is:
0.4 G
$\beta=\frac{\mu_0 I}{4 \pi r}\left(\sin \theta_1+\sin \theta_2\right)$
$\begin{aligned} & =\frac{10^{-7} \times 30}{12 \times 10^{-2}}\left(2 \times \frac{16}{20}\right) \\ & =4 \times 10^{-5} \mathrm{~T} \\ & =0.4 \times 10^{-4} \mathrm{~T} \\ & =0.4 \mathrm{G}\end{aligned}$
$\begin{aligned} & =\frac{10^{-7} \times 30}{12 \times 10^{-2}}\left(2 \times \frac{16}{20}\right) \\ & =4 \times 10^{-5} \mathrm{~T} \\ & =0.4 \times 10^{-4} \mathrm{~T} \\ & =0.4 \mathrm{G}\end{aligned}$
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.