A straight conductor of length 2m is carrying a current of 10 A. P is a paint at a distance of 23m from end A as shown in figure. then magnetic induction at P is

BP=μoi4πdsinθtanθ=223=13⇒θ=30o∴sinθ=12BP=10−7×1023×12T=143μT.

A straight conductor of length 2m is carrying a current of 10 A. P is a paint at a distance of 23m from end A as shown in figure. then magnetic induction at P is

BP=μoi4πdsinθtanθ=223=13⇒θ=30o∴sinθ=12BP=10−7×1023×12T=143μT.

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

A straight conductor of length 2m is carrying a current of 10 A. P is a paint at a distance of $2 \sqrt{3} m$ from end A as shown in figure. then magnetic induction at P is

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$4 \sqrt{3} μT$

$\frac{1}{4} μT$

$\frac{1}{4 \sqrt{3}} μT$

$2 \sqrt{3} μT$

answer is C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{l} B_{P} = \frac{μ_{o} i}{4 πd} sinθ \\ tanθ = \frac{2}{2 \sqrt{3}} = \frac{1}{\sqrt{3}} \Rightarrow θ = 30^{o} \\ ∴ sinθ = \frac{1}{2} \\ B_{P} = 10^{- 7} \times \frac{10}{2 \sqrt{3}} \times \frac{1}{2} T = \frac{1}{4 \sqrt{3}} μT . \end{array}$