In the figure shown a square loop PQRS of side ‘a’ and resistance ‘r’ is placed in near an infinitely long wire carrying a constant
current-I. The sides PQ and RS are parallel to the wire. The wire and the loop are in the same plane. The loop is rotated by 180°
about an axis parallel to the long wire and passing through the mid points of the side QR and PS. The total amount of charge
which passes through any point of the loop during rotation is

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

cannot be found because time of rotation is not given

$\frac{μ_{0} I a^{2}}{2 π r}$

$\frac{μ_{0} I a}{2 π r} \ln 2$

$\frac{μ_{0} I a}{π r} \ln 2$

answer is B.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

Magnetic flux density at a distance x from infinitely long current carrying wire $= B = \frac{μ I}{2 π x}$
flux linking the loop initially = $ϕ = \int_{a}^{2 a} Badx = \int_{a}^{2 a} \frac{μ I}{2 π x} adx = {|\frac{μ Ia}{2 π} lnx|}_{a}^{2 a} = \frac{μ Ia}{2 π} \ln 2$
Upon rotating the loop, total change in flux is $Δ ϕ = 2 \times \frac{μ Ia}{2 π} \ln 2 = \frac{μ Ia}{π} \ln 2$
emf induced in circuit $= \frac{Δ ϕ}{Δ t}$
total charge that has moved $= i Δ t = \frac{e m f}{r} Δ t = \frac{1 Δ ϕ}{r Δ t} Δ t = \frac{μ Ia}{π} \ln 2 \frac{1}{r}$