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Magnetics question

2021 · 1 Sep · Shift 2 · Q65
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  5. /2021 · 1 Sep · Shift 2 · Q65

Magnetics question

2021 · 1 Sep · Shift 2 · Q65

JEE MainPhysicsMagneticsMCQ+4 / −1
There are two infinitely long straight current carrying conductors and they are held at right angles to each other so that their common ends meet at the origin as shown in the figure given below. The ratio of current in both conductor is 1 : 1. The magnetic field at point P is ‾\underline{\hspace{2cm}}​. JEE Main 2021 (Online) 1st September Evening Shift Physics - Magnetic Effect of Current Question 114 English
  1. A
    μ0I4πxy[x2+y2+(x+y)]{{{\mu _0}I} \over {4\pi xy}}\left[ {\sqrt {{x^2} + {y^2}} + (x + y)} \right]4πxyμ0​I​[x2+y2​+(x+y)]
  2. B
    μ0I4πxy[x2+y2−(x+y)]{{{\mu _0}I} \over {4\pi xy}}\left[ {\sqrt {{x^2} + {y^2}} - (x + y)} \right]4πxyμ0​I​[x2+y2​−(x+y)]
  3. C
    μ0Ixy4π[x2+y2−(x+y)]{{{\mu _0}Ixy} \over {4\pi }}\left[ {\sqrt {{x^2} + {y^2}} - (x + y)} \right]4πμ0​Ixy​[x2+y2​−(x+y)]
  4. D
    μ0Ixy4π[x2+y2+(x+y)]{{{\mu _0}Ixy} \over {4\pi }}\left[ {\sqrt {{x^2} + {y^2}} + (x + y)} \right]4πμ0​Ixy​[x2+y2​+(x+y)]
View written solutionFree

Correct answer: A

  1. Interpret the geometry

    Two semi-infinite straight conductors meet at the origin and are mutually perpendicular.

    Let:

    • one conductor lie along the positive xxx-axis,
    • the other along the positive yyy-axis,
    • point PPP be at coordinates (x,y)(x,y)(x,y) in the first quadrant.

    Currents in both wires are equal: I1:I2=1:1I_1:I_2=1:1I1​:I2​=1:1, so each carries current III.

  2. Magnetic field due to the wire along the xxx-axis

    For a finite straight wire, the magnetic field at perpendicular distance rrr is

    B=μ0I4πr(sin⁡θ1+sin⁡θ2).B=\frac{\mu_0 I}{4\pi r}(\sin\theta_1+\sin\theta_2).B=4πrμ0​I​(sinθ1​+sinθ2​).

    For the semi-infinite wire along the xxx-axis:

    • perpendicular distance from P(x,y)P(x,y)P(x,y) to the xxx-axis is yyy,
    • one end is at infinity, so one angle is 90∘90^\circ90∘,
    • the other end is at the origin.

    The angle subtended by the segment from origin to point PPP gives

    sin⁡θ=xx2+y2.\sin\theta = \frac{x}{\sqrt{x^2+y^2}}.sinθ=x2+y2​x​.

    Therefore,

    Bx=μ0I4πy(1+xx2+y2).B_x=\frac{\mu_0 I}{4\pi y}\left(1+\frac{x}{\sqrt{x^2+y^2}}\right).Bx​=4πyμ0​I​(1+x2+y2​x​).

  3. Magnetic field due to the wire along the yyy-axis

    Similarly:

    • perpendicular distance from P(x,y)P(x,y)P(x,y) to the yyy-axis is xxx,
    • one angle is 90∘90^\circ90∘,
    • the other contributes
    sin⁡ϕ=yx2+y2.\sin\phi = \frac{y}{\sqrt{x^2+y^2}}.sinϕ=x2+y2​y​.

    Hence,

    By=μ0I4πx(1+yx2+y2).B_y=\frac{\mu_0 I}{4\pi x}\left(1+\frac{y}{\sqrt{x^2+y^2}}\right).By​=4πxμ0​I​(1+x2+y2​y​).

  4. Direction of the two fields

    By right-hand rule, at point PPP both magnetic fields are perpendicular to the plane and in the same direction, so magnitudes add:

    B=Bx+By.B=B_x+B_y.B=Bx​+By​.

  5. Add the two contributions

    B=μ0I4π[1y(1+xx2+y2)+1x(1+yx2+y2)].B=\frac{\mu_0 I}{4\pi}\left[\frac{1}{y}\left(1+\frac{x}{\sqrt{x^2+y^2}}\right)+\frac{1}{x}\left(1+\frac{y}{\sqrt{x^2+y^2}}\right)\right].B=4πμ0​I​[y1​(1+x2+y2​x​)+x1​(1+x2+y2​y​)].

    Expand:

    B=μ0I4π(1y+1x+xyx2+y2+yxx2+y2).B=\frac{\mu_0 I}{4\pi}\left(\frac{1}{y}+\frac{1}{x}+\frac{x}{y\sqrt{x^2+y^2}}+\frac{y}{x\sqrt{x^2+y^2}}\right).B=4πμ0​I​(y1​+x1​+yx2+y2​x​+xx2+y2​y​).

    Take common denominator xyxyxy:

    B=μ0I4πxy[x+y+x2+y2x2+y2].B=\frac{\mu_0 I}{4\pi xy}\left[x+y+\frac{x^2+y^2}{\sqrt{x^2+y^2}}\right].B=4πxyμ0​I​[x+y+x2+y2​x2+y2​].

    Since

    x2+y2x2+y2=x2+y2,\frac{x^2+y^2}{\sqrt{x^2+y^2}}=\sqrt{x^2+y^2},x2+y2​x2+y2​=x2+y2​,

    we get

    B=μ0I4πxy[x2+y2+(x+y)].B=\frac{\mu_0 I}{4\pi xy}\left[\sqrt{x^2+y^2}+(x+y)\right].B=4πxyμ0​I​[x2+y2​+(x+y)].

  6. Match with options

    This matches Option A:

    μ0I4πxy[x2+y2+(x+y)].\boxed{\frac{\mu_0 I}{4\pi xy}\left[\sqrt{x^2+y^2}+(x+y)\right]}.4πxyμ0​I​[x2+y2​+(x+y)]​.
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