An element of 0.05 hati m is placed at the origin as shown in figure which carries a large current of 10 A. The magnetic field at a distance of 1 m in perpendicular direction is

Question

An element of 0.05 hati m is placed at the origin as shown in figure which carries a large current of 10 A. The magnetic field at a distance of 1 m in perpendicular direction is <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/0ae7d71fd689bd8a47641ff5a5d280f7-.jpeg" /> (A) 4.5 × 10-8 T (B) 5.5 × 10-8 T (C) 5.0 × 10-8 T (D) 7.5 × 10-8 T

Tardigrade Admin · Accepted Answer

dB=(μ0/4π) (Idl sin θ/r2) Here, dl=Δ x=0.05 m, I=10 A, r=1 m sinθ=sin 90°=1, ∴ dB=10-7×(10×0.05×1/(1)2) =0.50×10-7 =5.0×10-8 T

Q. An element of $0.05 \hat{i} m$ is placed at the origin as shown in figure which carries a large current of $10 A$ . The magnetic field at a distance of $1 m$ in perpendicular direction is

$4.5 \times 1 0^{- 8} T$

$5.5 \times 1 0^{- 8} T$

$5.0 \times 1 0^{- 8} T$

$7.5 \times 1 0^{- 8} T$

$d B = \frac{μ _{0}}{4 π} \frac{I d l s in θ}{r ^{2}}$
Here, $d l = Δ x = 0.05 m, I = 10 A, r = 1 m$
$s in θ = s in 9 0^{\circ} = 1$ ,
$∴ d B = 1 0^{- 7} \times \frac{10 \times 0.05 \times 1}{( 1 ) ^{2}}$
$= 0.50 \times 1 0^{- 7}$
$= 5.0 \times 1 0^{- 8} T$

Q. An element of 0.05i^m is placed at the origin as shown in figure which carries a large current of 10A. The magnetic field at a distance of 1m in perpendicular direction is

4.5×10−8T

5.5×10−8T

5.0×10−8T

7.5×10−8T