The magnetic field due to a current carrying circular loop of radius 3 cm at a point on the axis at a distance of 4 cm from the centre is 54. What will be its value at the center of loop? (a) 125 (b) 150 (c) 250 (d) 75

Answer: (c). Bc = Baxis(R²+x²)3/2R³ R = 3 cm, x = 4 cm: (9+16)3/2 = 125, R³ = 27 Bc = 54 × 125/27 = 250 So the answer is (c) 250 μ T.

The magnetic field due to a current carrying circular loop of radius… | JEE Main 2004 (AIEEE)

Q108JEE Main 2004 · AIEEEMedium

Biot-Savart Law & Circular Coils

The magnetic field due to a current carrying circular loop of radius $3\text{ cm}$ at a point on the axis at a distance of $4\text{ cm}$ from the centre is $54\ \mu\text{T}.$ What will be its value at the center of loop?

Show answer and solution

Answer: (c)

$B_c = B_{axis}\dfrac{(R^2+x^2)^{3/2}}{R^3}$
$R = 3$ cm, $x = 4$ cm: $(9+16)^{3/2} = 125$ , $R^3 = 27$
$B_c = 54 \times \dfrac{125}{27} = 250\ \mu\text{T}$
So the answer is (c) $250\ \mu$ T.

Report a mistake

Biot-Savart Law & Circular Coils question from JEE Main 2004 (AIEEE)

More questions on Biot-Savart Law & Circular Coils