A conducting circular loop is placed in a uniform magnetic field of induction B tesla with its plane normal to the field. Now, radius of the loop starts shrinking at the rate (d r / d t). Then the induced e.m.f at the instant when the radius is r, is

Question

A conducting circular loop is placed in a uniform magnetic field of induction B tesla with its plane normal to the field. Now, radius of the loop starts shrinking at the rate (d r / d t). Then the induced e.m.f at the instant when the radius is r, is (A) π r B((d r/d t)) (B) 2 π r B((d r/d t)) (C) π r2 B((d r/d t)) (D) (π B r2/2)((d r/d t))

Tardigrade Admin · Accepted Answer

If the radius is r at a time t, then the instantaneous magnetic flux φ is given by φ=π r2 B Now, induced e.m.f. e is given by e=-(d φ/d t)=-(d/d t)(π r2 B) =-π B(2 r (d r/d t))=-2 π r B((d r/d t))

Q. A conducting circular loop is placed in a uniform magnetic field of induction $B$ tesla with its plane normal to the field. Now, radius of the loop starts shrinking at the rate $(d r / d t)$ . Then the induced e.m.f at the instant when the radius is $r$ , is

$π r B (\frac{d r}{d t})$

$2 π r B (\frac{d r}{d t})$

$π r^{2} B (\frac{d r}{d t})$

$\frac{π B r ^{2}}{2} (\frac{d r}{d t})$

If the radius is $r$ at a time $t$ , then the instantaneous magnetic flux $ϕ$ is given by $ϕ = π r^{2} B$
Now, induced e.m.f. $e$ is given by
$e^{=} - \frac{d ϕ}{d t} = - \frac{d}{d t} (π r^{2} B)$
$= - π B (2 r \frac{d r}{d t}) = - 2 π r B (\frac{d r}{d t})$

Q. A conducting circular loop is placed in a uniform magnetic field of induction B tesla with its plane normal to the field. Now, radius of the loop starts shrinking at the rate (dr/dt). Then the induced e.m.f at the instant when the radius is r, is

πrB(dtdr​)

2πrB(dtdr​)

πr2B(dtdr​)

2πBr2​(dtdr​)