The magnetic field B = 2t2 + 4t2 (where t = time) is applied perpendicular to the plane of a circular wire of radius r and resistance R. If all the units are in SI the electric charge that flows through the circular wire during t = 0 s to t = 2 s is

Question

The magnetic field B = 2t2 + 4t2 (where t = time) is applied perpendicular to the plane of a circular wire of radius r and resistance R. If all the units are in SI the electric charge that flows through the circular wire during t = 0 s to t = 2 s is (A) (6π r2/R) (B) (20π r2/R) (C) (32π r2/R) (D) (48π r2/R)

Tardigrade Admin · Accepted Answer

Given, B=2 t+4 t2 at t=0, B1=0 and at =2, B2=2 × 2+4(2)2 =4+16=20 Wb / m 2 We have, Δ Q =(Δ φ/R)=(π r2(B2-B1)/R) ∴ =(π r2[20-0]/R)=(20 π r2/R)

Q. The magnetic field $B = 2 t^{2} + 4 t^{2}$ (where t = time) is applied perpendicular to the plane of a circular wire of radius $r$ and resistance $R$ . If all the units are in $S I$ the electric charge that flows through the circular wire during $t = 0 s$ to $t = 2 s$ is

$\frac{6 π r ^{2}}{R}$

$\frac{20 π r ^{2}}{R}$

$\frac{32 π r ^{2}}{R}$

$\frac{48 π r ^{2}}{R}$

Given, $B = 2 t + 4 t^{2}$
at $t = 0, B_{1} = 0$
and at $= 2, B_{2} = 2 \times 2 + 4 (2)^{2}$
$= 4 + 16 = 20 Wb / m^{2}$
We have, $Δ Q = \frac{Δ ϕ}{R} = \frac{π r ^{2} ( B _{2} - B _{1} )}{R}$
$∴ = \frac{π r ^{2} [ 20 - 0 ]}{R} = \frac{20 π r ^{2}}{R}$

Q. The magnetic field B=2t2+4t2 (where t = time) is applied perpendicular to the plane of a circular wire of radius r and resistance R. If all the units are in SI the electric charge that flows through the circular wire during t=0s to t=2s is

R6πr2​

R20πr2​

R32πr2​

R48πr2​