Using Faraday's Law - AP Physics C Electricity & Magnetism

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Question

A wire loop of area is placed in a magnetic field whose magnitude varies with time according to the equation $B(t) = B_o sin(\omega t)$ . The loop is held in place, perpendicular to the field. If the wire has resistance , what is the magnitude of the maximum instantaneous current induced in the loop?

Answer

Relevant equations:

$|\epsilon| = \frac{d\phi}{dt} = \frac{d (B\cdot A)}{dt}$

$\phi = B \cdot A$

$\epsilon = IR$

Write the expression for the flux using the given expression for the magnetic field strength:

$\phi = B\cdot A = B_o sin(\omega t)A$

Take the derivative of this flux to find the induced emf:

$|\epsilon | = \frac{d\phi}{dt} = \frac{d (B_o sin(\omega t) A)}{dt} = B_o \omega cos(\omega t) A$

Determine the maximum value of $cos(\omega t)$ , and replace $cos(\omega t)$ with this value, to find the maximum induced emf:

$|\epsilon | _{max} = {B_o \omega * 1 * A} = B_o \omega A$

Plug this maximum induced emf into $\epsilon = IR$ , to find the maximum induced current:

$\epsilon = IR \rightarrow I = \frac{\epsilon}{R} = \frac{B_o \omega A}{R}$

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Question

A wire loop of area is placed in a magnetic field whose magnitude varies with time according to the equation $B(t) = B_o sin(\omega t)$ . The loop is held in place, perpendicular to the field. If the wire has resistance , what is the magnitude of the maximum instantaneous current induced in the loop?

Answer

Relevant equations:

$|\epsilon| = \frac{d\phi}{dt} = \frac{d (B\cdot A)}{dt}$

$\phi = B \cdot A$

$\epsilon = IR$

Write the expression for the flux using the given expression for the magnetic field strength:

$\phi = B\cdot A = B_o sin(\omega t)A$

Take the derivative of this flux to find the induced emf:

$|\epsilon | = \frac{d\phi}{dt} = \frac{d (B_o sin(\omega t) A)}{dt} = B_o \omega cos(\omega t) A$

Determine the maximum value of $cos(\omega t)$ , and replace $cos(\omega t)$ with this value, to find the maximum induced emf:

$|\epsilon | _{max} = {B_o \omega * 1 * A} = B_o \omega A$

Plug this maximum induced emf into $\epsilon = IR$ , to find the maximum induced current:

$\epsilon = IR \rightarrow I = \frac{\epsilon}{R} = \frac{B_o \omega A}{R}$

Compare your answer with the correct one above

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