Question

The figure below shows a 50 cm long bar attached to ideal frictionless parallel wires. The bar is free to slide along the wires but its orientation is fixed by the attachments. The bar is pulled by an external force at a constant velocity along the wires in the presence of a B = 2.60 T magnetic field which points into the screen. The parallel wires are connected to a R = 24.0 12 resistor and the bar makes an angle of Q = 30° with the wires. A I = 0.10 A induced current is measured in the resistor labeled R. The bar itself has a resistance r = 6.0 12 and a diameter of 0.6 cm x R x x X Part A - Determine the magnitude of the magnetic force on the bar. НА 0 ? Value Units Submit Request Answer

Answer 1

here,

length of bar , l = 50 cm = 0.5 m

B = 2.6 T

theta = 30 degree

a)

the magnitue of the magnetic force , Fb = B * I * L * sin(theta)

Fb = 2.6 * 0.1 * 0.5 * sin(30) N

Fb = 0.065 N

b)

as the electric flux is dercreasing with time

using Lenz's Law

the induced field must be into the page

using Right hand thumb rule

the direction of induced current is (CLockwise)

c)

the induced current , I = induced emf /(R + r)

I = B * v * l/(R + r)

0.1 = 2.6 * v * 0.5 /(24 + 5)

solving for v

v = 2.23 m/s

the speed of bar is 2.23 m/s

d)

r = p * l/(pi * (d/2)^2)

5 = p * 0.5 /(pi * (0.006/2)^2)

solving for p

p = 2.83 * 10^-4 ohm.m

the resistivity of bar is 2.83 * 10^-4 ohm.m