A 2.60-N steel bar, 0.850m long and having a resistance the 10.0?, rests horizontally on conducting wires connecting it to thecircuit presented in (Figure 1) . The bar is in a uniform, horizontal,1.60-T magnetic field and also is not attached come the wires in thecircuit.

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What is the acceleration of the bar simply after the switchS is closed?

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The ideas use to fix this difficulty are current department rule and also force on a existing carrying conductor in visibility of a magnetic field.

First calculate the identical resistance that the circuit by using collection and parallel mix of the circuit. After the calculate the total current in the circuit by making use of Ohm’s law. Following calculate the magnetic force experience top top the steel bar. Finally calculate the acceleration that the bar just after the move SSS is closed.


Current division rule, if present flow through more than one parallel paths, every of the parallel path shares a part of the present depending on the impedance that the path.

A existing carrying conductor when placed in presence of a magnetic ar experiences a magnetic force. The direction that magnetic force is provided by right hand dominion where as soon as all the fingers room pointed in the direction that magnetic ar with ignorance pointing in direction of current, the direction perpendicular to palm of best hand offers the direction the magnetic force.

From ohm’s law,

V=IRV = IRV=IR

Here, VVV is voltage, III is current, and also RRR is resistance.

The expression that magnetic force acting on the present carrying conductor, as placed in magnetic field,

Fm=BILsin⁡θF_\rmm = BIL\sin \theta Fm​=BILsinθ

Here, FmF_\rmmFm​ is magnetic force, BBB is magnetic field, III is current, LLL is length of conductor and also θ\theta θ is angle between current and magnetic field.

The expression of pressure in hatchet of mass and also acceleration.

F=maF = maF=ma

Here, mmm is mass, and aaa is acceleration.


The expression because that the present in an arm as soon as flowing v two parallel arms is offered by,

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Calculate the indistinguishable resistance of the circuit just after the switch is closed.

As the move is closed, the steel bar and also the 10Ω1\rm0 \Omega 10Ω resistor creates a parallel mix of resistors. The tantamount resistance that this parallel mix is,

ZP=(10×1010+10)Ω=5Ω\beginarrayc\\\rmZ_\rmP\rm = \left( \frac\rm10 \times 10\rm10 + 10 \right)\Omega \\\\ = \rm5 \Omega \\\endarrayZP​=(10+1010×10​)Ω=5Ω​

Here, ZP\rmZ_\rmPZP​ is identical resistance in parallel combination

Now calculate the tantamount resistance of the circuit.

The 25Ω25\rm \Omega 25Ω resistor and ZpZ_\rmpZp​ room in series so,

Zeq=25Ω+5Ω=30Ω\beginarrayc\\\rmZ_\rmeq = \rm25 \Omega + \rm5 \Omega \\\\ = 3\rm0 \Omega \\\endarrayZeq​=25Ω+5Ω=30Ω​

Now calculation the complete current in the circuit.

From ohm’s law,

V=IRV = IRV=IR

Here, VVV is voltage, III is current, and also RRR is resistance.

Rearrange the equation V=IRV = IRV=IR for existing III .

I=VRI = \fracVRI=RV​

Substitute 120.0V120.0\rm V120.0V for VVV and 30Ω3\rm0 \Omega 30Ω because that RRR .

I=120.0V30Ω=4.0A\beginarrayc\\I = \frac120.0\rm V3\rm0 \Omega \\\\ = 4.0\rm A\\\endarrayI=30Ω120.0V​=4.0A​

Calculate the current in the steel bar by making use of current division rule.

From the current department rule,

Im=R2R1+R2II_\rmm = \fracR_2R_1 + R_2IIm​=R1​+R2​R2​​I

Here, Im\rmI_\rmmIm​ is present through metal bar, III is total current with the circuit and also R1,R2R_1,R_2R1​,R2​ are the resistances in two parallel paths.

Substitute 4A4\rm A4A for III , 10Ω1\rm0 \Omega 10Ω for R2\rmR_\rm2R2​ and also 10Ω1\rm0 \Omega 10Ω for R1\rmR_1R1​ .

Im=(1010+10)4=2A\beginarrayc\\I_\rmm\rm = \left( \frac\rm10\rm10 + 10 \right)\rm4\\\\ = \rm2 A\\\endarrayIm​=(10+1010​)4=2A​

Calculate the force experience on the steel bar.

The expression that magnetic force acting on the present carrying conductor, as placed in magnetic field,

Fm=BILsin⁡θF_\rmm = BIL\sin \theta Fm​=BILsinθ

Here, FmF_mFm​ is magnetic force, BBB is magnetic field, III is current, LLL is length of conductor and

θ\theta θ is angle between current and magnetic field.

Substitute 1.60T1.60\rm T1.60T for, 2A\rm2 A2A because that I\rmII , 90∘90^\circ 90∘ because that θ\theta θ , and also 0.85m0.8\rm5 m0.85m because that LLL .

Fm=(1.60T)(2A)(0.85m)(sin⁡90∘)=(1.60T)(2A)(0.85m)(1)=2.72N\beginarrayc\\F_\rmm = \left( 1.60\rm T \right)\left( \rm2 A \right)\left( 0.8\rm5 m \right)\left( \sin 90^\circ \right)\\\\ = \left( 1.60\rm T \right)\left( \rm2 A \right)\left( 0.8\rm5 m \right)\left( 1 \right)\\\\ = 2.7\rm2 N\\\endarrayFm​=(1.60T)(2A)(0.85m)(sin90∘)=(1.60T)(2A)(0.85m)(1)=2.72N​

Calculate the mass of the steel bar.

The expression of load is,

w=mgw = mgw=mg

Here, www is weight, mmm is mass, and also ggg is acceleration as result of gravity.

Rearrange the equation w=mgw = mgw=mg because that mmm .

m=wgm = \fracwgm=gw​

Substitute 2.60N\rm2\rm.60 N2.60N because that www , and also 9.8m/s2\rm9\rm.8 m/\rms^29.8m/s2 because that ggg .

m=2.60N9.8m/s2=0.265kg\beginarrayc\\m = \frac\rm2\rm.60 N\rm9\rm.8 m/\rms^2\\\\ = 0.265\rm kg\\\endarraym=9.8m/s22.60N​=0.265kg​

Now calculation the acceleration the the metal bar.

The expression of force in term of mass and acceleration.

F=maF = maF=ma

Here, mmm is mass, and aaa is acceleration.

Rearrange the equation F=maF = maF=ma because that aaa .

a=Fma = \fracFma=mF​

Substitute 2.72N2.7\rm2 N2.72N because that FmF_\rmmFm​ and also 0.265kg0.265\rm kg0.265kg for mmm .

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a=2.72N0.265kg=10.26m/s2\beginarrayc\\a = \frac2.7\rm2 N0.265\rm kg\\\\ = 10.26\rm m/\rms^2\\\endarraya=0.265kg2.72N​=10.26m/s2​

Ans:

The acceleration of the steel bar is 10.26m/s210.26\rm m/\rms^210.26m/s2 .