- #1
Stephen Bulking
- 54
- 10
- Homework Statement
- This is not actually my homework since I already have the answers but the explanation is a little confusing, please check.
1. The resistance of the rectangular current loop is R. The metal rod is sliding to the right with the constant velocity of v. An infinite current I is placed near the loop(see figure). (please see word file attached)
The induced current around the loop:
answer is Counterclockwise , i=(μ0 vI×ln(b ∕ a))/2πR
3.Calculate the self-induced emf in the solenoid if the current it carries is decreasing at the rate of 50 A/s, know that the inductance of an air-core solenoid containing 300 turns and the length of the solenoid is 25 cm and its cross-sectional is 4 cm2:
A.-8.05mV B.5.30 mV C.13.05 mV D. 9.05mV
- Relevant Equations
- Lens' law
ε= -dΦ∕dt= -Ldi/dt
|ε|= dΦ∕dt
dΦ=Bds
ΔV=ε
i: induced current
L:inductant
1.
|ε|=dΦ/dt
B=μ0I×cos(0-cos180)/4πr
=μ0I/2πr
dΦ=BdS=μ0I×(vdtxdr)/2πr
ΔΦ=(μ0I×(vdt)/2π)x∫dr/r=(μ0I×(vdt)/2π)xln(b/a)
*the intergral goes from a to b
|ε|=dΦ/dt=(μ0I×v×ln(b/a)/2π)
i=ε/R=(μ0I×v×ln(b/a)/2πR)
2.
dI/dt=-50
B=μ0NI/L
ε=-dΦ/dt=-BdS/dt (*)
=-NBS/dt (wait what?)
=-N×N×I×S×μ0/dt
=-(N×N×S×μ0)×dI/dt (big what the hell)
= 9.05(mV)(*) my teacher wrote -∫BdS/dt but this is inconsistent with the given formula so I made this minor change
|ε|=dΦ/dt
B=μ0I×cos(0-cos180)/4πr
=μ0I/2πr
dΦ=BdS=μ0I×(vdtxdr)/2πr
ΔΦ=(μ0I×(vdt)/2π)x∫dr/r=(μ0I×(vdt)/2π)xln(b/a)
*the intergral goes from a to b
|ε|=dΦ/dt=(μ0I×v×ln(b/a)/2π)
i=ε/R=(μ0I×v×ln(b/a)/2πR)
2.
dI/dt=-50
B=μ0NI/L
ε=-dΦ/dt=-BdS/dt (*)
=-NBS/dt (wait what?)
=-N×N×I×S×μ0/dt
=-(N×N×S×μ0)×dI/dt (big what the hell)
= 9.05(mV)(*) my teacher wrote -∫BdS/dt but this is inconsistent with the given formula so I made this minor change
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