CAF123
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1. Homework Statement
A coil of wire with N turns is placed around the a long solenoid with n turns per unit length carrying a current I as shown below,
A)If a current I flows through the solenoid, derive an expression for the magnetic flux through the coil and explain why the size and shape of the coil is not important.
B)If the current in the solenoid is varied with time as given by the expression ##I(t) = I_o + I_1t + I_2t^2## and the resistance of the coil is R, derive an expression for the induced current in the external coil of wire.
2. Homework Equations
Ampere's Law
Magnetic Flux
3. The Attempt at a Solution
The magnetic field inside the solenoid is a constant ##\mu n I##. So the magnetic flux through the solenoid is ##\mu n I (\pi r^2).## Since we want the magnetic flux through the coil only this becomes ##\mu N I (\pi r^2)##(1).
However, the magnetic field caused by the current in the solenoid loop will induce a current inthe coil in the opposite direction to that of the loop in the solenoid.
By Ampere's Law, take a circle of radius smaller than that of the coil. The enclosed current is ## N I## and so ##\int B \cdot ds = \mu N I \Rightarrow B = \mu N I/(2 \pi R), R## the radius of the Amperian loop.
The coil (I assume) is perfectly symmetrical around the solenoid. So all B field contributions from the wire cancel. So B flux is given by (1). Is it correct?
If we take an Amperian loop to coincide with the radius of the solenoid, then no matter what shape the coil is, the enclosed current will always be the same. (But I don't see how this implies the B field would be same though). When they talk about size, do they mean a coil with a radius greater than that of solenoid?
2) Induced emf ,##E = \dot{\phi} = \mu N r^2 \pi [I_1 + 2 I_2 t],## so ##I_{ind} = E/R = \mu N r^2 \pi [I_1 + 2I_2 t]/R ##. (minus because current in opp direction)
Many thanks
A coil of wire with N turns is placed around the a long solenoid with n turns per unit length carrying a current I as shown below,
A)If a current I flows through the solenoid, derive an expression for the magnetic flux through the coil and explain why the size and shape of the coil is not important.
B)If the current in the solenoid is varied with time as given by the expression ##I(t) = I_o + I_1t + I_2t^2## and the resistance of the coil is R, derive an expression for the induced current in the external coil of wire.
2. Homework Equations
Ampere's Law
Magnetic Flux
3. The Attempt at a Solution
The magnetic field inside the solenoid is a constant ##\mu n I##. So the magnetic flux through the solenoid is ##\mu n I (\pi r^2).## Since we want the magnetic flux through the coil only this becomes ##\mu N I (\pi r^2)##(1).
However, the magnetic field caused by the current in the solenoid loop will induce a current inthe coil in the opposite direction to that of the loop in the solenoid.
By Ampere's Law, take a circle of radius smaller than that of the coil. The enclosed current is ## N I## and so ##\int B \cdot ds = \mu N I \Rightarrow B = \mu N I/(2 \pi R), R## the radius of the Amperian loop.
The coil (I assume) is perfectly symmetrical around the solenoid. So all B field contributions from the wire cancel. So B flux is given by (1). Is it correct?
If we take an Amperian loop to coincide with the radius of the solenoid, then no matter what shape the coil is, the enclosed current will always be the same. (But I don't see how this implies the B field would be same though). When they talk about size, do they mean a coil with a radius greater than that of solenoid?
2) Induced emf ,##E = \dot{\phi} = \mu N r^2 \pi [I_1 + 2 I_2 t],## so ##I_{ind} = E/R = \mu N r^2 \pi [I_1 + 2I_2 t]/R ##. (minus because current in opp direction)
Many thanks
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