Question

Shown in the figure is a circular loop of radius r and resistance R. A variable magnetic field of induction B = e^-t is established inside the coil. If the key (K) is closed at t = 0, the electrical power developed is equal to

• A. $\frac { \pi r ^ { 2 } } { R }$
• B. $\frac { 10 r ^ { 3 } } { R }$
• C. $\frac { \pi ^ { 2 } r ^ { 4 } R } { 5 }$
• D. $\frac { 10 r ^ { 4 } } { R }$

Answer 1

Answer: (D)

$\frac { 10 r ^ { 4 } } { R }$

Answer 2

The induced emf = ε = − $\frac { \mathrm { d } \phi } { \mathrm { dt } }$

= − = − A $\frac { \mathrm { dB } } { \mathrm { dt } }$ = − (πr²) $\frac { d } { d t } \left( e ^ { - t } \right) = \pi r ^ { 2 } e ^ { - t }$

⇒ ε_o = π r² $\left. e ^ { - t } \right| _ { t = 0 }$ = π r²

∴ The electrical power developed in the resistor just at the instant of closing the key = P = $\frac { \varepsilon _ { 0 } ^ { 2 } } { R } = \frac { \pi ^ { 2 } r ^ { 4 } } { R } \cong \frac { 10 r ^ { 4 } } { R }$

.Hence (D) is correct