An element $\Delta l=\Delta \mathrm{x} \hat{\mathrm{i}}$ is placed at the origin and carries a large current $\mathrm{I}=10 \mathrm{~A}$. The magnetic field on the y-axis at a distance of 0.5 m from the elements $\Delta \mathrm{x}$ of 1 cm length is :

(A) $4 \times 10^{-8} \mathrm{~T}$

(B) $8 \times 10^{-8} \mathrm{~T}$

(C) $12 \times 10^{-8} \mathrm{~T}$

(D) $10 \times 10^{-8} \mathrm{~T}$

JEE Main · 2024 · 06 Apr 2024 · Shift 1 · Magnetic Effects of Current

An element

Δ l = Δ x \hat{i}

is placed at the origin and carries a large current

I = 10 A

. The magnetic field on the y-axis at a distance of 0.5 m from the elements

Δ x

of 1 cm length is :

(A) $4 \times 1 0^{- 8} T$ ✓ Correct
(B) $8 \times 1 0^{- 8} T$
(C) $12 \times 1 0^{- 8} T$
(D) $10 \times 1 0^{- 8} T$

Answer: (A)

Solution

By the Biot-Savart law,

d B = \frac{μ _{0}}{4 π} \frac{I d l sin θ}{r ^{2}}

. The element is along

\hat{i}

and the point lies on the

y

-axis, so

θ = 9 0^{\circ}

sin θ = 1

, and

r = 0.5

d B = 1 0^{- 7} \times \frac{10 \times ( 1 \times 1 0 ^{- 2} )}{( 0.5 ) ^{2}} = 1 0^{- 7} \times \frac{0.1}{0.25} = 4 \times 1 0^{- 8} T .

JEE Main · 2024 · 06 Apr 2024 · Shift 1 · Single correct option

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