A conducting rod of length = 35.0 cm is free to slide on two parallel conducting

Question

A conducting rod of length = 35.0 cm is free to slide on two parallel conducting bars as shown in the figure below. Two resistors R1 = 2.00 and R2 = 5.00 are connected across the ends of the bars to form a loop. A constant magnetic field B = 2.60 T is directed perpendicularly into the page. An external agent pulls the rod to the left with a constant speed of v = 7.50 m/s.Find the following.

(a) the currents in both resistors

(b) the total power delivered to the resistance of the circuit

(c) the magnitude of the applied force that is needed to move the rod with this constant velocity

Explanation / Answer

Given that

Length of the rod (L) =35cm =0.35m

The resistance R1= 2.00 and R2 = 5.00

A constant magnetic field (B) = 2.60 T is directed perpendicularly into the page.

An external agent pulls the rod to the left with a constant speed of ( v) = 7.50 m/s.

The induced emf in the rod is given by   

e =BLv =(2.60)(0.35)(7.50)=6.825V

Here the resistors are connected in parallel then the net equivalent resistance is given by

R =R1R2/R1+R2 =2*5/(2+5) =10/7 =1.428ohm

Now the total current drawn in the circuit is I =e/R =6.825V/1.428ohm =4.775A

Therefore the current in each resisitors is given by

I2 =e/R1 =6.825/2 =3.4125A

I5 =e/R2 =6.825/5 =1.365A

B)

The power deliverd is given by

P =I2R =(4.775)2*(1.428) =32.593W

C)

The magnitude of the applied force is given by F =BIL =(2.60)(4.775)(0.35) =4.345N

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