Question

# A long solid cylindrical current carring conductor is placed along the zâˆ’ aixs, carrying current I in the negative z direction. The magnetic field â†’B at a point having coordinate (x,y) on the z=0 plane is

A
μ0I(y^ix^j)2π(x2+y2)
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B
μ0I(x^i+y^j)2π(x2+y2)
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C
μ0I(x^jy^j)2π(x2+y2)
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D
μ0I(x^iy^j)2π(x2+y2)
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Solution

## The correct option is A μ0I(y^i−x^j)2π(x2+y2)The plane representing z=0 is the x−y plane. Let us represent the current carrying conductor in x−y plane, carrying a current in −z direction as shown in the figure. The direction of magnetic field at point P will be tangential to the circular field line in clockwise direction with radius r. Here, →B=Bsinθ^i−Bcosθ^j ......(i) The magnitude of magnetic field due to solid cylinder at an outside point is, B=μ0I2πr Also, sinθ=yr and cosθ=xr Substituting these values in Eq. (i) →B=μ0I2πr[sinθ^i−cosθ^j] ⇒→B=μ0I2πr[yr^i−xr^j] or, →B=μ0I2πr2[y^i−x^j] From geometry of figure, r2=x2+y2 Thus, →B=μ0I[y^i−x^j]2π(x2+y2)

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