A 10-cm-long wire is pulled along a U-shaped conducting rail in a perpendicular magnetic field. The total resistance of the wire and rail is 0.20Ω. Pulling the wire with a force of 1.0 N causes 4.0 W of power to be dissipated in the circuit. a. What is the speed of the wire when pulled with a force of 1.0 N? b. What is the strength of the magnetic field?

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A 10-cm-long wire is pulled along a U-shaped conducting rail in a perpendicular magnetic field. The total resistance of the wire and rail is 0.20Ω. Pulling the wire with a force of 1.0 N causes 4.0 W of power to be dissipated in the circuit.  a. What is the speed of the wire when pulled with a force of 1.0 N?  b. What is the strength of the magnetic field?

Jeremy Merritt · Accepted Answer

Step 1  Given: length of wire (l)=10cm  Resistance of wire (R)=0.2Ω  Force (F)=1N  Power (P)=4W  a) Speed of wire.  we know that,  P=F.V  V=PF=41=4msec⁡  Step 2  b) Strength of the magnetic field (B)  P=(E.m.F)2R=(VBl)2R  P=V2B2l2R  4=42×B2×0.120.2  B=2.24T

Kindlein6h · Accepted Answer

In the above case we can say that power given by external agent to pull the rod must be equal to the power dissipated in the form of heat due to magnetic induction.   Part a)  when we pull the rod with constant speed then power required will be product of force and velocity here we will have    P=F.v  P=4W  v=4ms  now we will have  4=F⋅4  F=1N  So external force required will be 1 N   PART B)  now in order to find magnetic field strength we can say    P=v2B2L2R  here we know that induced EMF in the wire is E=vBL so power due to induced magnetic field is given by    P=E2R  4=42⋅B2⋅0.1020.20  by solving above equation we will have  B=2.24T

A 10-cm-long wire is pulled along a U-shaped conducting rail in a perp

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