A circular conducting loop of radius 29.0 cm is located in a region of homogeneous magnetic field...

Question:

A circular conducting loop of radius 29.0 cm is located in a region of homogeneous magnetic field of magnitude 0.300 T pointing perpendicular to the plane of the loop. The loop is connected in series with a resistor of 111 {eq}\Omega {/eq}. The magnetic field is now increased at a constant rate by a factor of 2.10 in 23.0s.

a) Calculate the magnitude of the current induced in the loop while the field is increasing.

b) With the magnetic field held constant at its new value of 0.63 T, calculate the magnitude of the average induced voltage in the loop while it is pulled horizontally out of the magnetic field region during a time interval of 3.70 s.

Induced emf:

Whenever a conductor, carrying current experiences a magnetic field, and due to variation in the magnetic field, the current is induced in another conductor, which induced an EMF. Induced emf can also be stated as the emf generated when there is no direct supply of energy.

Given data

• Radius of the circular loop is {eq}r = 29.0\;{\rm{cm}} {/eq}
• Magnitude of the magnetic field is {eq}B = 0.300\;{\rm{T}} {/eq}
• Resistance of the resistor is {eq}R = 111\;{\rm{\Omega }} {/eq}
• Factor of increasing is {eq}F = 2.10 {/eq}
• Time upto which field is increasing is {eq}23.0\;{\rm{s}} {/eq}

The expression for the induced emf is

{eq}e = \dfrac{{d\phi }}{{dt}} = \dfrac{{A\left( {{B_2} - {B_1}} \right)\cos \theta }}{t} {/eq}

The expression for the current is

{eq}I = \dfrac{e}{R} {/eq}

Substituting the values for the emf

{eq}\begin{align*} e &= \dfrac{{\left( {\pi {{\left( {0.29} \right)}^2}} \right)\left( {\left( {2.10 \times 0.300} \right) - 0.300} \right)\cos 0^\circ }}{{23}}\\ e &= 3.79 \times {10^{ - 3}}\;{\rm{V}} \end{align*} {/eq}

Substituting the values for the current

{eq}\begin{align*} I &= \dfrac{{3.79 \times {{10}^{ - 3}}}}{{111}}\\ I &= 3.41 \times {10^{ - 5}}\;{\rm{A}} \end{align*} {/eq}

Therefore, the induced current is {eq}3.41 \times {10^{ - 5}}\;{\rm{A}}.{/eq}

Substituting the values for the induced emf is

{eq}\begin{align*} e &= \dfrac{{\left( {\pi {{\left( {0.29} \right)}^2}} \right)\left( {0.63 - 0} \right)\cos 0^\circ }}{{3.70}}\\ e &= 0.044\;{\rm{V}} \end{align*} {/eq}

Therefore, the induced emf is {eq}0.044\;{\rm{V}} {/eq}