A)What is the magnitude of B? B)What is the value of Bx? C)What is the value of

Question

A)What is the magnitude of B?

B)What is the value of Bx?

C)What is the value of By?

D)What is the value of Bz?

E)Find the magnitude and direction of the magnetic field at P when the current in the left-hand wire is out of the page and the current in the right-hand wire is into the page. What is the magnitude of B?

F)What is the value of Bx?

G)What is the value of By?

H)What is the value of Bz?

Each of two long straight parallel wires 11 cm apart carries a current of 150 A. The figure shows a cross section, with the wires running perpendicular to the page and point P lying on the perpendicular bisector of the line between the wires. Find the magnitude and direction of the magnetic field at P when the current in the left-hand wire is out of the page and the current in the right-hand wire is also out of the page. A)What is the magnitude of B? B)What is the value of Bx? C)What is the value of By? D)What is the value of Bz? E)Find the magnitude and direction of the magnetic field at P when the current in the left-hand wire is out of the page and the current in the right-hand wire is into the page. What is the magnitude of B? F)What is the value of Bx? G)What is the value of By? H)What is the value of Bz?

Explanation / Answer

distance of point P= (d/(2)0.5) ) from both the wire
Applying Amperes Law
circle will be of radius (d/(2)0.5) )

B*2pi(d/(2)0.5)=uoI   
hence B=uoI /2pi(d/(2)0.5)    uo=4pi*10-7
by putting d=11cm=0.11m
we get B=3.856*10-4Tesla

By wire 1

B = B(-cos450 y+ sin450 z) y and z are unit vectors in y and z direction

By wire 2

B= B(-cos450 y - sin450 z)

Applying superposition principal
Total magnetic field at P= Sum of the fields due to each wire.

TotalMagnetic Field = -2Bcos45o y

Now putting B=uoI/2pi(d/(2)0/5) )=3.856*10-4Tesla we get total field=-5.45*10-4y Tesla

Hence
2)Bx=0,
3) By=-5.45*10-4y Tesla
4)Bz=0  

4)

Magnetic field due to wire 1 is same B = B(-cos450 y+ sin450 z)
but magnetic field due to wire 2 will be opposite of previous one
B= B(cos450 y + sin450 z) due to 2nd one

Hence total magnetic field = 2Bsin450 z=5.45*10-4z tesla

5)Bx=0,  
6) By=0,  
7)Bz=5.45*10-4tesla

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