Let $B_{P}$ , $B_{Q}$ and $B_{R}$ be the magnetic field produced by the three infinite wires P, Q and R respectively. The three wires are placed symmetrically inside an equilateral triangular loop as shown in figure. Currents in 3 wires are shown in figure.

If $\int_{A}^{B} {\vec{B}}_{P} \cdot \vec{d l} = 4 μ_{0} T - m$ and
$\int_{C}^{A} {\vec{B}}_{Q} \cdot \vec{d l} = - 15 μ_{0} T - m$ , then the value of ‘I’ (in ampere) 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

answer is 13.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

From the symmetry of figure, and using the fact that B $\propto$ I, we can find the line integral of $B_{p}, B_{Q} & B_{R}$ over various parts of closed loop. From given information.

$\int \vec{B} \cdot \vec{d l}$	$A \to B$	$B \to C$	$C \to A$
P	$4 μ_{0}$	$5 μ_{0}$	$4 μ_{0}$
Q	$- 12 μ_{0}$	$- 12 μ_{0}$	$- 15 μ_{0}$
R	$15 μ_{0}$	$12 μ_{0}$	$12 μ_{0}$

From Ampere’s circuital law,
$\oint \vec{B} \cdot \vec{d l} = \int_{A}^{B} \vec{B} \cdot \vec{d l} + \int_{B}^{C} \vec{B} \cdot \vec{d l} + \int_{C}^{A} \vec{B} \cdot \vec{d l} = μ_{0} (I - 3 I + 3 I) = μ_{0} I$

From the table,

$\oint \vec{B} \cdot \vec{d l} = \int_{A}^{B} \vec{B} \cdot \vec{d l} + \int_{B}^{C} \vec{B} \cdot \vec{d l} + \int_{C}^{A} \vec{B} \cdot \vec{d l} = μ_{0} \times 13$
$\Rightarrow$ I = 13 A