Question

Let $B_P,B_Q\text{and}{B}_R$ be the magnetic fields produced by the three infinite long wires P, Q and R, respectively. These wires are placed symmetrically inside an equilateral triangular loop as shown in the figure.

Current in 3 wires is as shown in figure below. If ${\int }_A^B{\overrightarrow{B}}_P.d\overrightarrow{l}=4{\mu }_0\text{and}{\int }_c^A{\overrightarrow{B}}_Q.d\overrightarrow{l}=-15{\mu }_0,$ considering SI units, the value of I is

• A. Information insufficient
• B. 15 A
• C. 5 A
• D. 13 A

Answer 1

The correct option is D 13 A

Given,
${\int }_A^B{\overrightarrow{B}}_P.d\overrightarrow{l}=4{\mu }_0$,

${\int }_C^A{\overrightarrow{B}}_Q.d\overrightarrow{l}=-15{\mu }_0$,

${\int }_B^C{\overrightarrow{B}}_P.d\overrightarrow{l}$ should be one third of

$|{\int }_C^A{\overrightarrow{B}}_Q.d\overrightarrow{l}|$ due to symmetry and current in Q is three times current in P

So,

$\Rightarrow {\int }_B^C{\overrightarrow{B}}_P.d\overrightarrow{l}=5{\mu }_0$

(+ve because direction of field by wire P and direction of circulation on loop are in same sense, both clockwise)

Similarly by symmetry (AB and AC are symmetric with respect to wire P)

$\Rightarrow {\int }_C^A{\overrightarrow{B}}_P.d\overrightarrow{l}={\int }_A^B{\overrightarrow{B}}_P.d\overrightarrow{l}=4{\mu }_0$

From, Ampere's law along the full loop
ABC,

$\oint {\overrightarrow{B}}_P.d\overrightarrow{l}={\mu }_0{i}_{en}$

So,
$\Rightarrow {\int }_A^B{\overrightarrow{B}}_P.d\overrightarrow{l}+{\int }_B^C{\overrightarrow{B}}_P.d\overrightarrow{l}+{\int }_C^A{\overrightarrow{B}}_P.d\overrightarrow{l}={\mu }_{0}i$

$\Rightarrow 4{\mu }_0+5{\mu }_0+4{\mu }_0={\mu }_{0}i$

(Substituting from the above equations)
$i=13A$

Hence, answer is (C).