Consider a closed loop ABCDA, in the form of a trapezium carrying current I. Match the following regarding the magnitude of magnetic field at point P.

Column – I Column – II
A) Magnetic field due to AB P) is greater than that due to DA
B) Magnetic field due to BC Q) is greater than that due to CD
C) Magnetic field due to DA R) is not equal to zero
D) Magnetic field due to complete figure S) is zero

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$A \to p; B \to s, q; C \to p, r; D \to q, r$

$A \to p, s; B \to q, s; C \to q, s, r; D \to p, s, r$

$A \to s; B \to s, q; C \to p, r; D \to p, q, s$

$A \to s; B \to p, q, r; C \to q, r D \to p, q, r$

answer is B.

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Detailed Solution

At P, magnetic field due to AB and CD will be zero.

${\vec{B}}_{B C} = \frac{μ_{0}}{4 π} \frac{I}{r} (\sin ϕ_{1} + \sin ϕ_{2}) (- \hat{k})$

${\vec{B}}_{D A} = \frac{μ_{0}}{4 π} \frac{I}{3 r} (\sin ϕ_{1} + \sin ϕ_{2}) (\hat{k})$

Net field at P:

$\vec{B} = {\vec{B}}_{B C} + {\vec{B}}_{D A} = \frac{2 μ_{0}}{4 π} \frac{I}{3 r} (\sin ϕ_{1} + \sin ϕ_{2}) (- \hat{k})$

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	Column – I		Column – II
A)	Magnetic field due to AB	P)	is greater than that due to DA
B)	Magnetic field due to BC	Q)	is greater than that due to CD
C)	Magnetic field due to DA	R)	is not equal to zero
D)	Magnetic field due to complete figure	S)	is zero