Question

Two resistors are connected in a circuit loop of area 5 m\(^2\), as shown in the figure below. The circuit loop is placed on the \( x-y \) plane. When a time-varying magnetic flux, with flux-density \( B(t) = 0.5t \) (in Tesla), is applied along the positive \( z \)-axis, the magnitude of current \( I \) (in Amperes, rounded off to two decimal places) in the loop is (answer in Amperes).

 

Hint:

When calculating the induced current, always consider the total resistance in the circuit and use the absolute value of the current since we are interested in the magnitude.

Answer 1

According to Faraday's Law of Induction, the induced electromotive force (EMF) in the loop is given by: \[ {EMF} = -\frac{d\Phi}{dt} \] where \( \Phi \) is the magnetic flux, defined as: \[ \Phi = B(t) \times A \] where:
\( B(t) = 0.5t \) (in Tesla),
\( A = 5 \, {m}^2 \) (area of the loop).
Thus, the flux \( \Phi \) is: \[ \Phi = 0.5t \times 5 = 2.5t \, {Weber} \] The induced EMF is: \[ {EMF} = -\frac{d}{dt}(2.5t) = -2.5 \, {V} \] Calculate the current using Ohm's law
The total resistance in the circuit is the sum of the resistances of the two resistors: \[ R_{{total}} = 2 \, \Omega + 2 \, \Omega = 4 \, \Omega \] Now, using Ohm's law \( I = \frac{{EMF}}{R} \), we calculate the current: \[ I = \frac{-2.5}{4} = -0.625 \, {A} \] The magnitude of the current is: \[ I = 0.62 \, {A} \] Thus, the correct answer is 0.62 A.