Question

A 3.10-N metal bar, 2.00 m long and having a resistance of 10.0 Ω, rests horizontally on conducting wires connecting it to the circuit shown in the figure (Figure 1) . The bar is in a uniform, horizontal, 1.40-T magnetic field and is not attached to the wires in the circuit.

What is the magnitude of the acceleration of the bar just after the switch S is closed?

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Answer 1

10 ohm resistance and 10 ohm bar are in parallel and their equivalent resistance is given as

R' = 10 x 10 / (10 + 10) = 5 ohm

R' and 25 ohm are in series and total resistance of the circuit is given as

R_total = R' + 25 = 5 + 25 = 30 ohm

total current coming from the battery is given as

i = V/R_total = 120/30 = 4 A

Voltage across the bar is given as

V_bar = V - Voltage across 25 ohm

V_bar = 120 - 4 x 25

V_bar = 20 volts

i_bar R_bar = 20                    i_bar = current flowing in bar    , R_bar = resistance of bar

i_bar = 20 / 10

i_bar = 2 A

using right hand rule , magnetic force Fb comes out to be upward and hence the bar is lifted up

force equation for the motion of bar is given as

F_b - F_g = ma                             f_g = force of gravity

i_bar B L - mg = ma

2 (1.4) (2) - 3.10 = (3.10/9.8) a

a = 7.9 m/s²