Question

A (20*20) cm² loop has a resistance of 0.10 Ω. A magnetic field perpendicular to the loop is B = 4t - 2t², where B is in tesla and t is in seconds.
What is the current in the loop at t = 0.0 s, t = 1.0 s, and t = 2.0 s?

Answer 1

Answer with Explanation:

We are given that

Area of loop=$(20\times 20) cm^2=400\times 10^{-4} m^2$

$1 cm^2=10^{-4} m^2$

Resistance, R=$0.1\Omega$

B=$4t-2t^2$

We know that magnetic flux

$\phi=BA$

Emf ,$E=\mid \frac{d\phi}{dt}\mid =\mid\frac{d(BA}{dt}\mid =\mid A\frac{dB}{dt}=400\times 10^{-4}\times \frac{4t-2t^2}{dt}\mid =\mid400\times 10^{-4}\times(4-4t)\mid$

Current, $I=\frac{E}{R}$

Current, $I=\frac{\mid 400\times 10^{-4}(4-4t)\mid }{0.1}=1.6\mid (1-t)\mid$

Substitute t=0 s

Then, I=$1.6\mid (1-0)\mid$=1.6 A

Substitute t=1 s

Then, I=$1.6\mid (1-1)\mid$=0

Substitute

t=2 s

Current, I=$1.6\mid(1-2)\mid$=1.6 A

Answer 2

Answer:

Explanation:

Area, A  = 20 x 20 cm² = 0.04 m²

Resistance, R = 0.10 ohm

Magnetic field, B = 4t - 2t²

the induced emf is given by

$e=\frac{d\phi }{dt}$

where, Ф i the flux linked with the coil.

$e=A\times \frac{dB }{dt}$

$e=0.04(4 - 4t)$

Induced current

i = e/ R

$i=\frac{0.04(4 - 4t)}{0.1}$

i = 0.4(4 - 4t)

For t = 0 s

i = 0.4 x 4 = 1.6 A

For t = 1 s

i = 0.4 ( 4 - 4) = 0 A

For t = 2 s

i = 0.4 ( 4 - 8) = 1.6 A in opposite direction