An element of $0.05^i m$ 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 10^{- 8} T

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5.5 \times 10^{- 8} T

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5.0 \times 10^{- 8} T

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7.5 \times 10^{- 8} T

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Solution

The correct option is C $5.0 \times 10^{- 8} T$
$d B = \frac{μ_{o}}{4 π} \frac{I d l s i n θ}{r^{2}}$

Here, $d l = Δ x = 0.05 m, I = 10 A, r = 1 m$

$s i n θ = s i n 90^{o} = 1,$

$d B = 10^{- 7} \times \frac{10 \times 0.05 \times 1}{(1)^{2}} = 0.50 \times 10^{- 7} = 5.0 \times 10^{- 8} T$

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An element of 0.05^im 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:

The correct option is C 5.0×10−8 TdB=μo4πIdlsinθr2Here, dl=Δx=0.05m,I=10A,r=1msinθ=sin90o=1,dB=10−7×10×0.05×1(1)2=0.50×10−7=5.0×10−8T

An element of $0.05^i m$ 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:

The correct option is C $5.0 \times 10^{- 8} T$
$d B = \frac{μ_{o}}{4 π} \frac{I d l s i n θ}{r^{2}}$

Here, $d l = Δ x = 0.05 m, I = 10 A, r = 1 m$

$s i n θ = s i n 90^{o} = 1,$

$d B = 10^{- 7} \times \frac{10 \times 0.05 \times 1}{(1)^{2}} = 0.50 \times 10^{- 7} = 5.0 \times 10^{- 8} T$