Two wires of infinite length are lying mutually at right angles to each others as shown in figure. Currents of 2A and 1.5A respectively are flowing in them. The value of magnetic induction at point p will be :

2 \times 10^{- 3} N / A m

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2 \times 10^{- 5} N / A m

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2 \times 10^{- 4} N / A m

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Solution

The correct option is A $2 \times 10^{- 5} N / A m$
$B = B_{1} + B_{2}$
$= \frac{μ_{0} i}{2 π x_{1}} + \frac{μ_{0} i}{2 π x_{2}}$
$= \frac{2 \times 10^{- 7} \times 2}{4 \times 10^{- 2}} + \frac{2 \times 10^{- 7} \times 1.5}{3 \times 10^{- 2}}$
$= 2 \times 10^{- 5} N / A - m$

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Similar questions

Q. Two insulated wires of infinite length are lying mutually at right angles to each other as shown in the figure. Current of 2 A and 1.5 A respectively are flowing in them. The value of magnetic induction at point P will be :

Q. Two concentric coils, each of radius equal to

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3 A

and

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, respectively, are flowing through the two coils. The magnetic induction, in

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[μ_{0} = 4 π \times 10^{- 7} Wb

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and

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are the currents flowing in each coil respectively. The magnetic induction in

W b / m^{2}

at the centre of the coils will be

(μ_{0} = 4 ϕ \times 10^{- 7} W b / A m)

Two similar coils of radius R are lying concentricaly with their planes at right angles to each other. The current flowing In them are l and 2l, respectively. The resultant magnetic field induction at the centre will be?

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Two wires of infinite length are lying mutually at right angles to each others as shown in figure. Currents of 2A and 1.5A respectively are flowing in them. The value of magnetic induction at point p will be :

The correct option is A 2×10−5N/AmB=B1+B2=μ0i2πx1+μ0i2πx2=2×10−7×24×10−2+2×10−7×1.53×10−2=2×10−5N/A−m

The correct option is A $2 \times 10^{- 5} N / A m$
$B = B_{1} + B_{2}$
$= \frac{μ_{0} i}{2 π x_{1}} + \frac{μ_{0} i}{2 π x_{2}}$
$= \frac{2 \times 10^{- 7} \times 2}{4 \times 10^{- 2}} + \frac{2 \times 10^{- 7} \times 1.5}{3 \times 10^{- 2}}$
$= 2 \times 10^{- 5} N / A - m$