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ivolga24 [154]
3 years ago
7

A straight vertical wire carries a current of 1.0 A into the page in a region where the magnetic field strength is 0.60 T. What

is the magnitude and direction of the magnetic force on a 1.0 cm section of this wire if the magnetic field direction is toward the east
Physics
1 answer:
Advocard [28]3 years ago
5 0

Answer:

force on the wire of section  cm will be 6\times 10^{-3}N

Direction of force on the wire will be in south direction        

Explanation:

We have given current in the wire i = 1 A

Magnetic field strength B = 0.6 T

We have to find the force on 1 cm section of the wire so l = 1 cm = 0.01 m

Force on the wire containing current is equal to

F=iBL

F=1\times 0.6\times 0.01=6\times 10^{-3}N

So force on the wire of section  cm will be 6\times 10^{-3}N

Direction of force on the wire will be in south direction

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You find a micrometer (a tool used to
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What is a nonzero net force called
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An older camera has a lens with a focal length of 60mm and uses 34-mm-wide film to record its images. Using this camera, a photo
lesya692 [45]

Answer:

24.71 mm

Explanation:

Distance is proportional to focal length, so

d∝f

which means

\frac{d'_1}{d'_2}=\frac{f_1}{f_2}

Magnification of first lens

M_2=-\frac{d'_1}{d_1}

                   and

M_2=\frac{h'_1}{h_1}

Similarly, magnification of second lens

M_2=-\frac{d'_2}{d_1}

                   and

M_2=\frac{h'_2}{h_1}

From the above equations we get

\frac{M_1}{M_2}=\frac{d'_1}{d_2'}

                   and

\frac{M_1}{M_2}=\frac{h'_1}{h_2'}

which means,

\frac{d'_1}{d_2'}=\frac{h'_1}{h_2'}

and

\frac{d'_1}{d_2'}=\frac{f_1}{f_2}

So, we get

\frac{f_1}{f_2}=\frac{h'_1}{h_2'}\\\Rightarrow f_2=f_1\times\frac{h_2'}{h'_1}\\\Rightarrow f_2=60\times\frac{14}{34}=24.71\ mm

∴ Focal length should this camera's lens is 24.71 mm

6 0
3 years ago
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