Question

Solve previous year Question of iit jee
chapter :- magnetic effect of current​

Answer 1

Answer:

option a is correct answer

Hope it will help you

Answer 2

Answer:

Option (a)

Explanation:

The magnetic field at a point on the acid of a circular loop at a distance x from the centre is

B = μ₀IR² / 2(R²+x²)³ˡ²

Given , B = 54μΤ , x = 4cm , R = 3cm

putting the given values in eq (i) we get

• 54 = μ₀I × (3)²/ 2(3² + 4²) ³ˡ²

• → 54 = 9μ₀I / 2 (25)³ˡ² = 9μ₀I / 2× (5)³

• ∴ μ₀I = 54× 2× 125 /9

• → μ₀ I = 1500μΤ -cm .--------(i)

Now, Putting x = 0 in eq (i) , Magnetic fiels at the centre of loop is

• B = μ₀IR² / 2R³

• =μ₀I / 2R = 1500 /2×3 [from eq (I) ]

= 250μΤ Answer