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MCQMediumJEE 2026Biot–Savart Law

JEE Physics 2026 Question with Solution

The current passing through a conducting loop in the form of equilateral triangle of side 43 cm4\sqrt{3} \, \text{cm} is 2 A2 \, \text{A}. The magnetic field at its centroid is α×10−5 T\alpha \times 10^{-5} \, \text{T}. The value of α\alpha is _____. (Given : μ0=4π×10−7\mu_0 = 4\pi \times 10^{-7} SI units)

  • A

    333\sqrt{3}

  • B

    232\sqrt{3}

  • C

    3\sqrt{3}

  • D

    32\frac{\sqrt{3}}{2}

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: Side of the equilateral triangle is a=43 cma = 4\sqrt{3} \, \text{cm} and current is I=2 AI = 2 \, \text{A}.

Find: The value of α\alpha in the magnetic field B=α×10−5 TB = \alpha \times 10^{-5} \, \text{T}.

The distance from the centroid to a side of an equilateral triangle is

d=a23d = \frac{a}{2\sqrt{3}}

Substituting a=43 cma = 4\sqrt{3} \, \text{cm},

d=4323=2 cm=0.02 md = \frac{4\sqrt{3}}{2\sqrt{3}} = 2 \, \text{cm} = 0.02 \, \text{m}

The magnetic field due to one side at the centroid is

B1=μ0I4πd(sin⁡θ1+sin⁡θ2)B_1 = \frac{\mu_0 I}{4\pi d}(\sin \theta_1 + \sin \theta_2)

For an equilateral triangle, the two angles are 60∘60^\circ and 60∘60^\circ.

Therefore,

B1=4π×10−7×24π×0.02(2sin⁡60∘)B_1 = \frac{4\pi \times 10^{-7} \times 2}{4\pi \times 0.02}(2\sin 60^\circ)

Using sin⁡60∘=32\sin 60^\circ = \frac{\sqrt{3}}{2},

B1=2×10−70.02(3)=3×10−5 TB_1 = \frac{2 \times 10^{-7}}{0.02}(\sqrt{3}) = \sqrt{3} \times 10^{-5} \, \text{T}

The total magnetic field at the centroid is three times the field due to one side:

Bnet=3B1=33×10−5 TB_{net} = 3B_1 = 3\sqrt{3} \times 10^{-5} \, \text{T}

Comparing with α×10−5 T\alpha \times 10^{-5} \, \text{T},

α=33\alpha = 3\sqrt{3}

Therefore, the correct option is A.

Regular Polygon Formula

Given: A conducting loop is in the form of an equilateral triangle, so it is a regular polygon with n=3n = 3 sides.

Find: The magnetic field at the centroid and hence the value of α\alpha.

Use the regular polygon result mentioned in the hint:

B=nμ0I2πRtan⁡(πn)B = \frac{n\mu_0 I}{2\pi R} \tan\left(\frac{\pi}{n}\right)

where RR is the circumradius.

For an equilateral triangle of side a=43 cma = 4\sqrt{3} \, \text{cm}, the circumradius is

R=a3=4 cm=0.04 mR = \frac{a}{\sqrt{3}} = 4 \, \text{cm} = 0.04 \, \text{m}

Now substitute n=3n = 3, I=2 AI = 2 \, \text{A}, and tan⁡(π/3)=3\tan(\pi/3) = \sqrt{3}:

B=3⋅4π×10−7⋅22π⋅0.04⋅3=33×10−5 TB = \frac{3 \cdot 4\pi \times 10^{-7} \cdot 2}{2\pi \cdot 0.04} \cdot \sqrt{3} = 3\sqrt{3} \times 10^{-5} \, \text{T}

Hence, α=33\alpha = 3\sqrt{3}, so the correct option is A. This shortcut works because the centroid of an equilateral triangle is also its circumcenter.

Common mistakes

  • Using the side length directly in the straight-wire formula is incorrect because the formula requires the perpendicular distance from the point to the wire. Here that distance is d=a23d = \frac{a}{2\sqrt{3}}, not aa.

  • Taking the field of only one side and forgetting to multiply by 33 is wrong because all three sides produce magnetic fields in the same direction at the centroid. Add the contributions of all three sides.

  • Using the wrong angle in sin⁡θ\sin \theta leads to an incorrect result. At the centroid, each end of a side subtends 60∘60^\circ in the given straight-wire expression, so use sin⁡60∘\sin 60^\circ for both terms.

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