A conducting circular loop of area 2.5× 10- 3 m2 and resistance 1 Ω is placed perpendicular to a uniform time varying magnetic field B(t)=0.6sin(50 π t) T . What is the net charge (in mC ) flowing through the loop during t=0 and t = 10 textms ?

Question

A conducting circular loop of area 2.5× 10- 3 m2 and resistance 1 Ω is placed perpendicular to a uniform time varying magnetic field B(t)=0.6sin(50 π t) T . What is the net charge (in mC ) flowing through the loop during t=0 and t = 10 textms ? (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Net charge =(Δ φ/R)=(A (B2 - B1)/R) =(2 . 5 × (10)- 3)(.0.6.)[sin (50 π (. 10 .)/(10)3) - sin 0] =1.5× 10- 3[sin (π /2) - sin 0] =1.5× 10- 3C=1.5mC

Q. A conducting circular loop of area $2.5 \times 1 0^{- 3} m^{2}$ and resistance $1 Ω$ is placed perpendicular to a uniform time varying magnetic field $B (t) = 0.6 s in (50 π t) T$ . What is the net charge (in $m C$ ) flowing through the loop during $t = 0$ and $t = 10 ms$ ?

Net charge $= \frac{Δ ϕ}{R} = \frac{A ( B _{2} - B _{1} )}{R}$
$= (2.5 \times (10)^{- 3}) (0.6) [s in \frac{50 π ( 10 )}{( 10 ) ^{3}} - s in 0]$
$= 1.5 \times 1 0^{- 3} [s in \frac{π}{2} - s in 0]$
$= 1.5 \times 1 0^{- 3} C = 1.5 m C$

Q. A conducting circular loop of area 2.5×10−3m2 and resistance 1Ω is placed perpendicular to a uniform time varying magnetic field B(t)=0.6sin(50πt)T . What is the net charge (in mC ) flowing through the loop during t=0 and t=10ms ?

Net charge =RΔϕ​=RA(B2​−B1​)​ =(2.5×(10)−3)(0.6)[sin(10)350π(10)​−sin0] =1.5×10−3[sin2π​−sin0] =1.5×10−3C=1.5mC

Q. A conducting circular loop of area $2.5 \times 1 0^{- 3} m^{2}$ and resistance $1 Ω$ is placed perpendicular to a uniform time varying magnetic field $B (t) = 0.6 s in (50 π t) T$ . What is the net charge (in $m C$ ) flowing through the loop during $t = 0$ and $t = 10 ms$ ?

Net charge $= \frac{Δ ϕ}{R} = \frac{A ( B _{2} - B _{1} )}{R}$
$= (2.5 \times (10)^{- 3}) (0.6) [s in \frac{50 π ( 10 )}{( 10 ) ^{3}} - s in 0]$
$= 1.5 \times 1 0^{- 3} [s in \frac{π}{2} - s in 0]$
$= 1.5 \times 1 0^{- 3} C = 1.5 m C$