Understanding Electromagnetic Induction - AP Physics C: Electricity and Magnetism

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Question

A metal ring is placed in a uniform magnetic field perpendicular to the plane of the ring. An emf of magnitude 15V is induced around the ring by increasing the field through it from zero to 5mT at a constant rate. If the area enclosed by the ring is $1 $*10^{-2}$$m^2$$ , what is the time interval over which the field is increased?

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Answer

Relevant equations:

$\epsilon = $\frac{-\Delta \Phi}{\Delta t}$ = \frac {- \Delta (B\cdot A)}{\Delta t}$

Given:

$A = 1 * $10^{-2}$ $m^2$$

$\Delta B = 5mT = 5 * $10^{-3}$T$

$\epsilon = 15 V$

Plug in (disregarding direction of emf):

$15 V= $\frac{(5 * $10^{-3}$$T)(1 * $10^{-2}$$m^2$)}{\Delta t}$

$\Delta t = $$\frac{(510^{-3}$$T)(1 $10^{-2}$$m^2$)}{15V}= 3.33 * $10^{-6}$s = 3.33\mu s$

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Question

A metal ring is placed in a uniform magnetic field perpendicular to the plane of the ring. An emf of magnitude 15V is induced around the ring by increasing the field through it from zero to 5mT at a constant rate. If the area enclosed by the ring is $1 $*10^{-2}$$m^2$$ , what is the time interval over which the field is increased?

Tap to reveal answer

Answer

Relevant equations:

$\epsilon = $\frac{-\Delta \Phi}{\Delta t}$ = \frac {- \Delta (B\cdot A)}{\Delta t}$

Given:

$A = 1 * $10^{-2}$ $m^2$$

$\Delta B = 5mT = 5 * $10^{-3}$T$

$\epsilon = 15 V$

Plug in (disregarding direction of emf):

$15 V= $\frac{(5 * $10^{-3}$$T)(1 * $10^{-2}$$m^2$)}{\Delta t}$

$\Delta t = $$\frac{(510^{-3}$$T)(1 $10^{-2}$$m^2$)}{15V}= 3.33 * $10^{-6}$s = 3.33\mu s$

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Question

A metal ring is placed in a uniform magnetic field perpendicular to the plane of the ring. An emf of magnitude 15V is induced around the ring by increasing the field through it from zero to 5mT at a constant rate. If the area enclosed by the ring is $1 $*10^{-2}$$m^2$$ , what is the time interval over which the field is increased?

Tap to reveal answer

Answer

Relevant equations:

$\epsilon = $\frac{-\Delta \Phi}{\Delta t}$ = \frac {- \Delta (B\cdot A)}{\Delta t}$

Given:

$A = 1 * $10^{-2}$ $m^2$$

$\Delta B = 5mT = 5 * $10^{-3}$T$

$\epsilon = 15 V$

Plug in (disregarding direction of emf):

$15 V= $\frac{(5 * $10^{-3}$$T)(1 * $10^{-2}$$m^2$)}{\Delta t}$

$\Delta t = $$\frac{(510^{-3}$$T)(1 $10^{-2}$$m^2$)}{15V}= 3.33 * $10^{-6}$s = 3.33\mu s$

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Question

A metal ring is placed in a uniform magnetic field perpendicular to the plane of the ring. An emf of magnitude 15V is induced around the ring by increasing the field through it from zero to 5mT at a constant rate. If the area enclosed by the ring is $1 $*10^{-2}$$m^2$$ , what is the time interval over which the field is increased?

Tap to reveal answer

Answer

Relevant equations:

$\epsilon = $\frac{-\Delta \Phi}{\Delta t}$ = \frac {- \Delta (B\cdot A)}{\Delta t}$

Given:

$A = 1 * $10^{-2}$ $m^2$$

$\Delta B = 5mT = 5 * $10^{-3}$T$

$\epsilon = 15 V$

Plug in (disregarding direction of emf):

$15 V= $\frac{(5 * $10^{-3}$$T)(1 * $10^{-2}$$m^2$)}{\Delta t}$

$\Delta t = $$\frac{(510^{-3}$$T)(1 $10^{-2}$$m^2$)}{15V}= 3.33 * $10^{-6}$s = 3.33\mu s$

← Didn't Know|Knew It →

Question

A metal ring is placed in a uniform magnetic field perpendicular to the plane of the ring. An emf of magnitude 15V is induced around the ring by increasing the field through it from zero to 5mT at a constant rate. If the area enclosed by the ring is $1 $*10^{-2}$$m^2$$ , what is the time interval over which the field is increased?

Tap to reveal answer

Answer

Relevant equations:

$\epsilon = $\frac{-\Delta \Phi}{\Delta t}$ = \frac {- \Delta (B\cdot A)}{\Delta t}$

Given:

$A = 1 * $10^{-2}$ $m^2$$

$\Delta B = 5mT = 5 * $10^{-3}$T$

$\epsilon = 15 V$

Plug in (disregarding direction of emf):

$15 V= $\frac{(5 * $10^{-3}$$T)(1 * $10^{-2}$$m^2$)}{\Delta t}$

$\Delta t = $$\frac{(510^{-3}$$T)(1 $10^{-2}$$m^2$)}{15V}= 3.33 * $10^{-6}$s = 3.33\mu s$

← Didn't Know|Knew It →

Question

A metal ring is placed in a uniform magnetic field perpendicular to the plane of the ring. An emf of magnitude 15V is induced around the ring by increasing the field through it from zero to 5mT at a constant rate. If the area enclosed by the ring is $1 $*10^{-2}$$m^2$$ , what is the time interval over which the field is increased?

Tap to reveal answer

Answer

Relevant equations:

$\epsilon = $\frac{-\Delta \Phi}{\Delta t}$ = \frac {- \Delta (B\cdot A)}{\Delta t}$

Given:

$A = 1 * $10^{-2}$ $m^2$$

$\Delta B = 5mT = 5 * $10^{-3}$T$

$\epsilon = 15 V$

Plug in (disregarding direction of emf):

$15 V= $\frac{(5 * $10^{-3}$$T)(1 * $10^{-2}$$m^2$)}{\Delta t}$

$\Delta t = $$\frac{(510^{-3}$$T)(1 $10^{-2}$$m^2$)}{15V}= 3.33 * $10^{-6}$s = 3.33\mu s$

← Didn't Know|Knew It →

Question

A metal ring is placed in a uniform magnetic field perpendicular to the plane of the ring. An emf of magnitude 15V is induced around the ring by increasing the field through it from zero to 5mT at a constant rate. If the area enclosed by the ring is $1 $*10^{-2}$$m^2$$ , what is the time interval over which the field is increased?

Tap to reveal answer

Answer

Relevant equations:

$\epsilon = $\frac{-\Delta \Phi}{\Delta t}$ = \frac {- \Delta (B\cdot A)}{\Delta t}$

Given:

$A = 1 * $10^{-2}$ $m^2$$

$\Delta B = 5mT = 5 * $10^{-3}$T$

$\epsilon = 15 V$

Plug in (disregarding direction of emf):

$15 V= $\frac{(5 * $10^{-3}$$T)(1 * $10^{-2}$$m^2$)}{\Delta t}$

$\Delta t = $$\frac{(510^{-3}$$T)(1 $10^{-2}$$m^2$)}{15V}= 3.33 * $10^{-6}$s = 3.33\mu s$

← Didn't Know|Knew It →

Question

A metal ring is placed in a uniform magnetic field perpendicular to the plane of the ring. An emf of magnitude 15V is induced around the ring by increasing the field through it from zero to 5mT at a constant rate. If the area enclosed by the ring is $1 $*10^{-2}$$m^2$$ , what is the time interval over which the field is increased?

Tap to reveal answer

Answer

Relevant equations:

$\epsilon = $\frac{-\Delta \Phi}{\Delta t}$ = \frac {- \Delta (B\cdot A)}{\Delta t}$

Given:

$A = 1 * $10^{-2}$ $m^2$$

$\Delta B = 5mT = 5 * $10^{-3}$T$

$\epsilon = 15 V$

Plug in (disregarding direction of emf):

$15 V= $\frac{(5 * $10^{-3}$$T)(1 * $10^{-2}$$m^2$)}{\Delta t}$

$\Delta t = $$\frac{(510^{-3}$$T)(1 $10^{-2}$$m^2$)}{15V}= 3.33 * $10^{-6}$s = 3.33\mu s$

← Didn't Know|Knew It →

Question

A metal ring is placed in a uniform magnetic field perpendicular to the plane of the ring. An emf of magnitude 15V is induced around the ring by increasing the field through it from zero to 5mT at a constant rate. If the area enclosed by the ring is $1 $*10^{-2}$$m^2$$ , what is the time interval over which the field is increased?

Tap to reveal answer

Answer

Relevant equations:

$\epsilon = $\frac{-\Delta \Phi}{\Delta t}$ = \frac {- \Delta (B\cdot A)}{\Delta t}$

Given:

$A = 1 * $10^{-2}$ $m^2$$

$\Delta B = 5mT = 5 * $10^{-3}$T$

$\epsilon = 15 V$

Plug in (disregarding direction of emf):

$15 V= $\frac{(5 * $10^{-3}$$T)(1 * $10^{-2}$$m^2$)}{\Delta t}$

$\Delta t = $$\frac{(510^{-3}$$T)(1 $10^{-2}$$m^2$)}{15V}= 3.33 * $10^{-6}$s = 3.33\mu s$

← Didn't Know|Knew It →

Question

A metal ring is placed in a uniform magnetic field perpendicular to the plane of the ring. An emf of magnitude 15V is induced around the ring by increasing the field through it from zero to 5mT at a constant rate. If the area enclosed by the ring is $1 $*10^{-2}$$m^2$$ , what is the time interval over which the field is increased?

Tap to reveal answer

Answer

Relevant equations:

$\epsilon = $\frac{-\Delta \Phi}{\Delta t}$ = \frac {- \Delta (B\cdot A)}{\Delta t}$

Given:

$A = 1 * $10^{-2}$ $m^2$$

$\Delta B = 5mT = 5 * $10^{-3}$T$

$\epsilon = 15 V$

Plug in (disregarding direction of emf):

$15 V= $\frac{(5 * $10^{-3}$$T)(1 * $10^{-2}$$m^2$)}{\Delta t}$

$\Delta t = $$\frac{(510^{-3}$$T)(1 $10^{-2}$$m^2$)}{15V}= 3.33 * $10^{-6}$s = 3.33\mu s$

← Didn't Know|Knew It →

Question

A metal ring is placed in a uniform magnetic field perpendicular to the plane of the ring. An emf of magnitude 15V is induced around the ring by increasing the field through it from zero to 5mT at a constant rate. If the area enclosed by the ring is $1 $*10^{-2}$$m^2$$ , what is the time interval over which the field is increased?

Tap to reveal answer

Answer

Relevant equations:

$\epsilon = $\frac{-\Delta \Phi}{\Delta t}$ = \frac {- \Delta (B\cdot A)}{\Delta t}$

Given:

$A = 1 * $10^{-2}$ $m^2$$

$\Delta B = 5mT = 5 * $10^{-3}$T$

$\epsilon = 15 V$

Plug in (disregarding direction of emf):

$15 V= $\frac{(5 * $10^{-3}$$T)(1 * $10^{-2}$$m^2$)}{\Delta t}$

$\Delta t = $$\frac{(510^{-3}$$T)(1 $10^{-2}$$m^2$)}{15V}= 3.33 * $10^{-6}$s = 3.33\mu s$

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Knew It

Understanding Electromagnetic Induction - AP Physics C: Electricity and Magnetism

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