Question

A 10.0 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 0.330 ohms . Pulling the wire at a steady speed of 4.00 m/s causes 4.40 W of power to be dissipated in the circuit.a) How big is the pulling force?b) What is the strength of the magnetic field?

Answer 1

Answer

given,

length of wire = 10 cm = 0.1 m

resistance of the wire =  0.330 ohms

speed of pulling = 4 m/s

Power = 4.40 W

a) Force of pull = ?

 $P = F_{pull} v$

 $F_{pull} =\dfrac{P}{v}$

 $F_{pull} =\dfrac{4.40}{4}$

 $F_{pull} =1.1\ N$

b) using formula

  $P = \dfrac{B^2l^2v^2}{R}$

  where B is the magnetic field

   v is the pulling velocity

   R is the resistance of the wire

  $B =\sqrt{\dfrac{PR}{l^2v^2}}$

  $B =\sqrt{\dfrac{F_{pull} \times v R}{l^2v^2}}$

  $B =\sqrt{\dfrac{F_{pull} \times R}{l^2v}}$

  $B =\sqrt{\dfrac{1.1 \times 0.33}{0.1^2\times 4}}$

  $B =\sqrt{9.075}$

      B = 3.01 T