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Question: Answered & Verified by Expert

Match List I with List II

List – I
(Current configuration)
List – II
(Magnetic field at point
O)
A I. B0=μ0I4πr[π+2]
B II. B0=μ04Ir
C III. B0=μ0I2πr[π-1]
D IV. B0=μ0I4πr[π+1]

Choose the correct answer from the option given below:

PhysicsMagnetic Effects of CurrentJEE MainJEE Main 2023 (25 Jan Shift 1)
Options:
  • A A-III, B-IV, C-I, D-II
  • B A-I, B-III, C-IV, D-II
  • C A-III, B-I, C-IV, D-II
  • D A-II, B-I, C-IV, D-III
Solution:
2798 Upvotes Verified Answer
The correct answer is: A-III, B-I, C-IV, D-II

(A)

Here net magnetic field will be sum of magnetic field due to straight wires ab, de and due to loop bcd.

So, Bab=μ04πIr=Bde (directed out of the plane in both cases)

Bbcd=μ04πIr(2π) (in the plane)

Thus, magnetic field at O is

BO=-μ04πIr+μ04πIr(2π)-μ04πIr

BO=μ02πIr(π-1)  (III)

(B)

Here net magnetic field will be sum of magnetic field due to straight wires ab, de and semicircular arc bcd.

Bab=μ04πIr=Bde (directed out of the plane in both cases)

Bbcd=μ04πIr(π) (out of the plane)

Thus, net magnetic field at O is

BO=μ04πIr+μ04πIr(π)+μ04πIr

B0=μ04πIr(π+2) (I)

(C)

Here net magnetic field will be sum of magnetic field due to straight wires ab, de and due to loop bcd.

Bab=μ04πIr (directed in the plane)

Bbcd=μ04πIr(π) (in the plane)

Bde=0 (passing through the axis)

Thus, magnetic field at O is

BO=μ04πIr(π+1)  (IV)

(D)

Here net magnetic field will be sum of magnetic field due to straight wires ab, de and due to loop bcd.

Bab=0=Bde (passing through the axis)

Bbcd=μ04πIr(π) (out of the plane)

Thus, magnetic field at O is

BO=μ0I4r (II).

A-III, B-I, C-IV, D-II

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