Understanding Electromagnetic Induction - AP Physics C: Electricity and Magnetism

Card 0 of 11

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?

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):

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

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

Compare your answer with the correct one above

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?

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):

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

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

Compare your answer with the correct one above

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?

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):

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

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

Compare your answer with the correct one above

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?

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):

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

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

Compare your answer with the correct one above

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?

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):

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

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

Compare your answer with the correct one above

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?

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):

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

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

Compare your answer with the correct one above

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?

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):

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

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

Compare your answer with the correct one above

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?

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):

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

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

Compare your answer with the correct one above

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?

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):

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

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

Compare your answer with the correct one above

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?

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):

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

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

Compare your answer with the correct one above

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?

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):

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

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

Compare your answer with the correct one above

Tap the card to reveal the answer

Understanding Electromagnetic Induction - AP Physics C: Electricity and Magnetism

Relevant equations: Given: Plug in (disregarding direction of emf):

Relevant equations: Given: Plug in (disregarding direction of emf):

Relevant equations: Given: Plug in (disregarding direction of emf):

Relevant equations: Given: Plug in (disregarding direction of emf):

Relevant equations: Given: Plug in (disregarding direction of emf):

Relevant equations: Given: Plug in (disregarding direction of emf):

Relevant equations: Given: Plug in (disregarding direction of emf):

Relevant equations: Given: Plug in (disregarding direction of emf):

Relevant equations: Given: Plug in (disregarding direction of emf):

Relevant equations: Given: Plug in (disregarding direction of emf):

Relevant equations: Given: Plug in (disregarding direction of emf):