Wires 1 and 2 carrying currents i1 and i2 respectively are incl-Turito

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Wires 1 and 2 carrying currents i1 and i2 respectively are inclined at an angle q to each other What is the force on a small element dl of wire 2 at a distance of r from wire 1 (as shown in figure) due to the magnetic field of wire 1

Physics-General

$\frac{μ_{0}}{2 π r} i_{1} i_{2} d l t a n θ$
$\frac{μ_{0}}{2 π r} i_{1} i_{2} d l s i n θ$
$\frac{μ_{0}}{2 π r_{1} i_{2} d l c o s θ}$
$\frac{μ_{0}}{4 π r} i_{1} i_{2} d l s i n θ$

Answer:The correct answer is: $fraction numerator mu subscript 0 end subscript over denominator 2 pi r subscript 1 end subscript i subscript 2 end subscript d l c o s invisible function application theta end fraction$

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A particle $A$ is projected from the ground with an initial velocity of $10 blank m s to the power of negative 1 end exponent$ at an angle of 60 $degree$ with horizontal. From what height $h$ should an another particle $B$ be projected horizontally with velocity $5 m s to the power of negative 1 end exponent$ so that both the particles collide in ground at point $C$ if both are projected simultaneously? $left parenthesis g equals 10 blank m s to the power of negative 2 end exponent right parenthesis$

Horizontal component of velocity of

A

is 10

c o s blank 6 0 degree

5 blank m s to the power of negative 1 end exponent

which is equal to the velocity of

B

in horizontal direction. They will collide at

C

if time of flight of the particles are equal or

t subscript A end subscript equals t subscript B end subscript

fraction numerator 2 u sin invisible function application theta over denominator g end fraction equals square root of fraction numerator 2 h over denominator g end fraction end root open parentheses therefore h equals fraction numerator 1 over denominator 2 end fraction g t subscript B end subscript superscript 2 end superscript close parentheses

h equals fraction numerator 2 u to the power of 2 end exponent sin to the power of 2 end exponent invisible function application theta over denominator g end fraction

equals fraction numerator 2 open parentheses 10 close parentheses to the power of 2 end exponent open parentheses fraction numerator square root of 3 over denominator 2 end fraction close parentheses to the power of 2 end exponent over denominator 10 end fraction equals 15 blank m

A particle $A$ is projected from the ground with an initial velocity of $10 blank m s to the power of negative 1 end exponent$ at an angle of 60 $degree$ with horizontal. From what height $h$ should an another particle $B$ be projected horizontally with velocity $5 m s to the power of negative 1 end exponent$ so that both the particles collide in ground at point $C$ if both are projected simultaneously? $left parenthesis g equals 10 blank m s to the power of negative 2 end exponent right parenthesis$

physics-General

Horizontal component of velocity of

A

is 10

c o s blank 6 0 degree

5 blank m s to the power of negative 1 end exponent

which is equal to the velocity of

B

in horizontal direction. They will collide at

C

if time of flight of the particles are equal or

t subscript A end subscript equals t subscript B end subscript

fraction numerator 2 u sin invisible function application theta over denominator g end fraction equals square root of fraction numerator 2 h over denominator g end fraction end root open parentheses therefore h equals fraction numerator 1 over denominator 2 end fraction g t subscript B end subscript superscript 2 end superscript close parentheses

h equals fraction numerator 2 u to the power of 2 end exponent sin to the power of 2 end exponent invisible function application theta over denominator g end fraction

equals fraction numerator 2 open parentheses 10 close parentheses to the power of 2 end exponent open parentheses fraction numerator square root of 3 over denominator 2 end fraction close parentheses to the power of 2 end exponent over denominator 10 end fraction equals 15 blank m

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V subscript r end subscript equals 2 omega R sin invisible function application left parenthesis omega t right parenthesis

t equals T divided by 2 comma blank V subscript r end subscript equals 0

So two half cycles will take place

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physics-General

So,

V subscript r end subscript equals 2 omega R sin invisible function application left parenthesis omega t right parenthesis

t equals T divided by 2 comma blank V subscript r end subscript equals 0

So two half cycles will take place

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Let f : R →R and g : R →R be continuous functions, then the value of $\int_{- π / 2}^{π /2}$ (f(x) + f(-x)) (g(x) – g(-x)) dx, is equal to

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Wires 1 and 2 carrying currents i1 and i2 respectively are inclined at an angle q to each other What is the force on a small element dl of wire 2 at a distance of r from wire 1 (as shown in figure) due to the magnetic field of wire 1

$\frac{μ_{0}}{2 π r} i_{1} i_{2} d l t a n θ$
$\frac{μ_{0}}{2 π r} i_{1} i_{2} d l s i n θ$
$\frac{μ_{0}}{2 π r_{1} i_{2} d l c o s θ}$
$\frac{μ_{0}}{4 π r} i_{1} i_{2} d l s i n θ$

Answer:The correct answer is: $fraction numerator mu subscript 0 end subscript over denominator 2 pi r subscript 1 end subscript i subscript 2 end subscript d l c o s invisible function application theta end fraction$

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Wires 1 and 2 carrying currents i1 and i2 respectively are inclined at an angle q to each other What is the force on a small element dl of wire 2 at a distance of r from wire 1 (as shown in figure) due to the magnetic field of wire 1

Answer:The correct answer is:

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Answer:The correct answer is: $fraction numerator mu subscript 0 end subscript over denominator 2 pi r subscript 1 end subscript i subscript 2 end subscript d l c o s invisible function application theta end fraction$

Two long parallel conductors carry currents $i subscript 1 end subscript equals 3 A text and end text i subscript 2 end subscript equals 3 A$ both are directed into the plane of paper The magnitude of resultant magnetic field at point ‘P; is

Two long parallel conductors carry currents $i subscript 1 end subscript equals 3 A text and end text i subscript 2 end subscript equals 3 A$ both are directed into the plane of paper The magnitude of resultant magnetic field at point ‘P; is

Two wires AO and OC carry currents i as shown in figure One end of both the wires extends to infinity $angle A O C equals alpha$ , the magnitude of magnetic field at a point P on the bisector of the two wires at a distance r from O is

Two wires AO and OC carry currents i as shown in figure One end of both the wires extends to infinity $angle A O C equals alpha$ , the magnitude of magnetic field at a point P on the bisector of the two wires at a distance r from O is

Let f : R →R and g : R →R be continuous functions, then the value of $\int_{- π / 2}^{π /2}$ (f(x) + f(-x)) (g(x) – g(-x)) dx, is equal to

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