MHT CET · Physics · Magnetic Effects of Current

Six very long insulated copper wires are bound together to form a cable. The currents carried by the wires are $l_1=+10 A, l_2=-13 A, l_3=+10 A, l_4=+7 A,$ $l_5=-12 A$ and $l_6=+18 A$. The magnetic induction at a perpendicular distance of 10 cm from the cable is $\left(\mu_0=4 \pi \times \frac{10^{-7} Wb}{ A }- m \right)$

A $40 μ T$
B $37.5 μ T$
C $30 μ T$
D $35 μ T$

Verified Solution

Answer & Solution

Correct Answer

(A) $40 μ T$

Step-by-step Solution

Detailed explanation

Net current due to all wires,

i_{n e t} = i_{1} + i_{2} + i_{3} + i_{4} + i_{5} + i_{6}

i_{n e t} = 10 + 13 + 10 + 7 - 12 + 18 = 20 A

We know, magnetic field due to an infinitely long straight conductor at a perpendicular distance r from it is given by

B = \frac{μ_{0} i}{2 π r} = \frac{μ_{0} i_{n e t}}{2 π r}

where,

i

= current in wire
and

r

= perpendicular distance.

= \frac{4 π \times 10^{- 7} \times 20}{2 π \times 10 \times 10^{- 2}} = 4 \times 10^{- 5}

B = 40 μ T

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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