Question

Previous year Jee mains question

chapter :- magnetic effect of current . ​

Answer 1

Answer:

We already know that Magnetic field due to any arc having current 'I' at an angle 'θ' and radius 'r' is μ₀Iθ/4πr

Now,

For arc(AD) :-

The magnetic field (B1) due to arc(AD) at the origin O is in upward direction from the plane.

$θ = \frac{\pi}{6}$

r = OA = a

→ B1 = (μ₀Iπ/6)/4πa

= μ₀I/24a

For arc(BC) :-

The magnetic field (B2) due to arc(AD) at the origin O is inside the plane

$θ = \frac{\pi}{6}$

r = OC = b

→B2 = (μ₀Iπ/6)/4πb

= μ₀I/24b

Hence,

The magnetic field due to the loop ABCD at the origin = B1-B2

$= \frac{μ₀I }{24a} - \frac{μ₀I }{24b}$

$= \frac{μ₀I }{24} ( \frac{1}{a} - \frac{1}{b} )$

$= \frac{μ₀I }{24} ( \frac{b - a}{ab} )$

Answer 2

Answer:

Option (b) μ₀I(b-a)/ 24ba

Explanation:

Given :- A ABCD loop structure with angle 30° at the origin.

Find :- Net magnetic field due to loop ABCD at O .

So for this we have to find Bnet = B(AB) + B(BC) + B(CD) + B (OA)

Let's say B(AB) = B₁ , B(BC) = B₂ , B(CD) = B₃ , B(DA) = B₄

• Bnet = B₁ + B₂ + B₃ + B₄

But , Here B₁ and B₃ cancelled each other's magnetic field because they are in the opposite direction (A/q right hand thumb rule )

Now , We have Bnet = B₂ + B₄

magnetic fiel of B₂ = μ₀Iθ / 4πR

where θ = π/6 and R = a

• ∴ B₂ = μ₀I/4πr × π/6 = μ₀I/24b/a

Magnetic field of B₄ = (μ₀I θ/4πR]

Here, θ = π/6

Here, θ = π/6 Radius "R" = b

[∵ B₄ is Opposite direction from B₂ ]

• ∴ B₄ = - μ₀I/24b

Bnet = B₂ + B₄ = μ₀I/24a +(-μ₀I/24b)

Bnet = μ₀I/24 (1/a - 1/b)

• Bnet = μ₀I (b-a)/24ab Answer