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Magnetism question

2021 · Shift 2 · Q44
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Magnetism question

2021 · Shift 2 · Q44

JEE AdvancedPhysicsMagnetismMultiple correct+4 / −2
Two concentric circular loops, one of radius R and the other of radius 2R, lie in the xy-plane with the origin as their common center, as shown in the figure. The smaller loop carries current I1 in the anti-clockwise direction and the larger loop carries current I2 in the clockwise direction, with I2 > 2I1. B→\overrightarrow BB(x, y) denotes the magnetic field at a point (x, y) in the xy-plane. Which of the following statement(s) is(are) correct? JEE Advanced 2021 Paper 2 Online Physics - Magnetism Question 30 English
  1. A
    B→\overrightarrow BB(x, y) is perpendicular to the xy-plane at any point in the plane
  2. B
    | B→\overrightarrow BB(x, y) | depends on x and y only through the radial distance r=x2+y2r = \sqrt {{x^2} + {y^2}}r=x2+y2​
  3. C
    | B→\overrightarrow BB(x, y) | is non-zero at all points for r < R
  4. D
    B→\overrightarrow BB(x, y) points normally outward from the xy-plane for all the points between the two loops
View written solutionFree

Correct answer: A, B

Analysis of the Problem

We have two concentric circular current loops in the xy-plane, centered at the origin.

  • Loop 1 (smaller): Radius RRR, current I1I_1I1​ (anti-clockwise).
  • Loop 2 (larger): Radius 2R2R2R, current I2I_2I2​ (clockwise).
  • Given condition: I2>2I1I_2 > 2I_1I2​>2I1​.
  • The magnetic field at a point (x,y)(x, y)(x,y) in the xy-plane is denoted by B→(x,y)\overrightarrow B(x, y)B(x,y).

We will evaluate each statement systematically.

Evaluation of Option A

Statement A: B→(x,y)\overrightarrow B(x, y)B(x,y) is perpendicular to the xy-plane at any point in the plane.

  1. Biot-Savart Law: The magnetic field element dB→d\overrightarrow BdB produced by a current element Idl→I d\overrightarrow lIdl at a position r→′\overrightarrow r'r′ is given by: dB→=μ04πIdl→×(r→−r→′)∣r→−r→′∣3d\overrightarrow B = \frac{{\mu_0}}{{4\pi}} \frac{{I d\overrightarrow l \times (\overrightarrow r - \overrightarrow r')}}{{|\overrightarrow r - \overrightarrow r'|^3}}dB=4πμ0​​∣r−r′∣3Idl×(r−r′)​ where r→\overrightarrow rr is the position vector of the point where the field is being calculated.

  2. Applying to the setup:

    • Both current loops lie in the xy-plane. This means any current element dl→d\overrightarrow ldl for either loop is a vector in the xy-plane.
    • The point of observation (x,y)(x, y)(x,y) is also in the xy-plane, so its position vector r→=xi^+yj^\overrightarrow r = x\hat i + y\hat jr=xi^+yj^​ is in the xy-plane.
    • The position vector of the current element r→′\overrightarrow r'r′ is also in the xy-plane.
    • Therefore, the vector difference (r→−r→′)(\overrightarrow r - \overrightarrow r')(r−r′) is also a vector in the xy-plane.
  3. Cross Product: The cross product of two vectors in the xy-plane, dl→d\overrightarrow ldl and (r→−r→′)(\overrightarrow r - \overrightarrow r')(r−r′), will result in a vector perpendicular to the xy-plane, i.e., a vector along the z-axis (either +k^+\hat k+k^ or −k^-\hat k−k^). dl→×(r→−r→′)=(dB→)zk^d\overrightarrow l \times (\overrightarrow r - \overrightarrow r') = (d\overrightarrow B)_z \hat kdl×(r−r′)=(dB)z​k^

  4. Total Field: The total magnetic field B→\overrightarrow BB is the vector sum (integral) of all such elements dB→d\overrightarrow BdB from both loops. Since every dB→d\overrightarrow BdB is directed along the z-axis, their sum B→\overrightarrow BB must also be directed along the z-axis. B→(x,y)=Bz(x,y)k^\overrightarrow B(x, y) = B_z(x, y) \hat kB(x,y)=Bz​(x,y)k^ This means the magnetic field is perpendicular to the xy-plane at any point in the plane.

  5. Conclusion for A: Statement A is correct.

Evaluation of Option B

Statement B: |B→\overrightarrow BB(x, y)| depends on x and y only through the radial distance r=x2+y2r = \sqrt {{x^2} + {y^2}}r=x2+y2​.

  1. Symmetry: The physical system consists of two concentric circular loops centered at the origin. This configuration has cylindrical symmetry about the z-axis.

  2. Effect of Rotation: If we rotate the coordinate system (or the observation point) around the z-axis by any angle, the current distribution looks exactly the same. The physics of the situation is unchanged.

  3. Field Magnitude: Consequently, the magnitude of the magnetic field, |B→\overrightarrow BB|, cannot depend on the azimuthal angle ϕ\phiϕ. It can only depend on the distance from the axis of symmetry (the z-axis), which is the radial distance r=x2+y2r = \sqrt{x^2 + y^2}r=x2+y2​, and the coordinate along the axis, zzz. Since we are in the xy-plane, z=0z=0z=0.

  4. Conclusion for B: Therefore, |B→\overrightarrow BB(x, y)| depends on x and y only through rrr. Statement B is correct.

Evaluation of Option C

Statement C: |B→\overrightarrow BB(x, y)| is non-zero at all points for r<Rr < Rr<R.

  1. Field Direction: For r<Rr < Rr<R, the point is inside both loops.

    • The field due to loop 1 (I1I_1I1​, anti-clockwise) points in the +z direction (out of the plane). Let's call its z-component B1zB_{1z}B1z​.
    • The field due to loop 2 (I2I_2I2​, clockwise) points in the -z direction (into the plane). Let's call its z-component B2zB_{2z}B2z​ (which will be a negative value).
    • The total field's z-component is Bz(r)=∣B1(r)∣−∣B2(r)∣B_z(r) = |B_1(r)| - |B_2(r)|Bz​(r)=∣B1​(r)∣−∣B2​(r)∣. (Using magnitudes of fields)
  2. Field at the Center (r=0):

    • ∣B1(0)∣=μ0I12R|B_1(0)| = \frac{{\mu_0 I_1}}{{2R}}∣B1​(0)∣=2Rμ0​I1​​
    • ∣B2(0)∣=μ0I22(2R)=μ0I24R|B_2(0)| = \frac{{\mu_0 I_2}}{{2(2R)}} = \frac{{\mu_0 I_2}}{{4R}}∣B2​(0)∣=2(2R)μ0​I2​​=4Rμ0​I2​​
    • The net field at the center is Bz(0)=μ0I12R−μ0I24R=μ04R(2I1−I2)B_z(0) = \frac{{\mu_0 I_1}}{{2R}} - \frac{{\mu_0 I_2}}{{4R}} = \frac{{\mu_0}}{{4R}}(2I_1 - I_2)Bz​(0)=2Rμ0​I1​​−4Rμ0​I2​​=4Rμ0​​(2I1​−I2​).
    • Given I2>2I1I_2 > 2I_1I2​>2I1​, the term (2I1−I2)(2I_1 - I_2)(2I1​−I2​) is negative. So, Bz(0)<0B_z(0) < 0Bz​(0)<0. The field at the center is non-zero.
  3. Field near the inner loop (r → R⁻):

    • As the observation point approaches the wire of the smaller loop (r→Rr \to Rr→R from the inside), the magnetic field contribution from this loop, B1B_1B1​, becomes very large and tends to infinity (∣B1(r)∣→∞|B_1(r)| \to \infty∣B1​(r)∣→∞). The direction is +z.
    • The contribution from the larger loop, B2B_2B2​, remains finite at r=Rr=Rr=R.
    • Therefore, as r→R−r \to R^-r→R−, the net field Bz(r)=∣B1(r)∣−∣B2(r)∣B_z(r) = |B_1(r)| - |B_2(r)|Bz​(r)=∣B1​(r)∣−∣B2​(r)∣ tends to +∞+\infty+∞.
  4. Intermediate Value Theorem: The function Bz(r)B_z(r)Bz​(r) is continuous for 0≤r<R0 \le r < R0≤r<R. We have found that Bz(0)B_z(0)Bz​(0) is negative and Bz(r)B_z(r)Bz​(r) becomes positive and large as r→R−r \to R^-r→R−. By the Intermediate Value Theorem, there must exist a radial distance r0r_0r0​ such that 0<r0<R0 < r_0 < R0<r0​<R where Bz(r0)=0B_z(r_0) = 0Bz​(r0​)=0.

  5. Conclusion for C: The magnetic field is zero at some points for r<Rr<Rr<R. Therefore, statement C is incorrect.

Evaluation of Option D

Statement D: B→\overrightarrow BB(x, y) points normally outward from the xy-plane for all the points between the two loops.

  1. Region of Interest: This statement concerns the region R<r<2RR < r < 2RR<r<2R.

  2. Field Directions:

    • For a point outside the smaller loop (r>Rr > Rr>R), the field B→1\overrightarrow B_1B1​ due to the anti-clockwise current I1I_1I1​ points into the plane, i.e., in the -z direction.
    • For a point inside the larger loop (r<2Rr < 2Rr<2R), the field B→2\overrightarrow B_2B2​ due to the clockwise current I2I_2I2​ also points into the plane, i.e., in the -z direction.
  3. Total Field: Since both B→1\overrightarrow B_1B1​ and B→2\overrightarrow B_2B2​ point in the -z direction in the region R<r<2RR < r < 2RR<r<2R, their vector sum B→=B→1+B→2\overrightarrow B = \overrightarrow B_1 + \overrightarrow B_2B=B1​+B2​ must also point in the -z direction.

  4. Conclusion for D: The field points normally inward, not outward. Therefore, statement D is incorrect.

Final Summary

  • Statement A is correct.
  • Statement B is correct.
  • Statement C is incorrect.
  • Statement D is incorrect.

The correct statements are A and B.

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