JEE Main 2022 · 30 Jul · Shift 2 · Physics Q16

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PhysicsSingle correct+4 / -1Question 16 of 90

In the following loop, magnitude of the magnetic field produced at the centre (O) of the loop is : (Assume AB and CD are infinitely extended):

A semi-infinite straight wire A→B along the horizontal line through centre O, then three-quarters of a circle of radius R about O from B round to C directly below O, then a semi-infinite straight wire C→D running back leftward a distance R below O.

\dfrac{\mu_0I}{4\pi R}\left(\dfrac32\pi\right)

\dfrac{\mu_0I}{2\pi R}

\dfrac{\mu_0I}{4\pi R}\left(\dfrac32\pi+1\right)

\dfrac{\mu_0I}{4\pi R}\left(\dfrac32\pi-1\right)

Answer and worked solution

Official answer: C

Three-quarter loop gives $\frac34\cdot\frac{\mu_0I}{2R}=\frac{\mu_0I}{4\pi R}\cdot\frac{3\pi}2$ . AB lies on the line through O: no field. CD is semi-infinite, ending at the foot of the perpendicular from O at distance R: $\frac{\mu_0I}{4\pi R}$ , in the same sense.

Total $\dfrac{\mu_0I}{4\pi R}\left(\dfrac{3\pi}2+1\right)$ .

(C)

Topics:Biot-Savart law

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JEE Main 2022 · 30 Jul · Shift 2, Physics question 16