Physics
General
Easy

Question

Rohit completes a semi circular path of radius R in 10 seconds. Calculate average speed and average velocity in ms-1.

  1. fraction numerator 2 πR over denominator 10 end fraction comma fraction numerator 2 R over denominator 10 end fraction
  2. πR over 10 comma R over 10
  3. fraction numerator negative R over denominator 10 end fraction comma fraction numerator 2 R over denominator 10 end fraction
  4. fraction numerator 2 πR over denominator 10 end fraction comma R over 10

The correct answer is: fraction numerator negative R over denominator 10 end fraction comma fraction numerator 2 R over denominator 10 end fraction

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Related Questions to study

General
physics-

If Young’s modulus of elasticity Y for a material is one and half times its rigidity coefficient eta comma the Poisson’s ratio sigma will be

Young’s modulus, Y equals fraction numerator 3 ɳ over denominator 2 end fraction
We know that Y equals 2 ɳ open parentheses 1 plus sigma close parentheses
therefore blank fraction numerator 3 ɳ over denominator 2 end fraction equals 2 ɳ open parentheses 1 plus sigma close parentheses
blank sigma equals negative fraction numerator 1 over denominator 4 end fraction

If Young’s modulus of elasticity Y for a material is one and half times its rigidity coefficient eta comma the Poisson’s ratio sigma will be

physics-General
Young’s modulus, Y equals fraction numerator 3 ɳ over denominator 2 end fraction
We know that Y equals 2 ɳ open parentheses 1 plus sigma close parentheses
therefore blank fraction numerator 3 ɳ over denominator 2 end fraction equals 2 ɳ open parentheses 1 plus sigma close parentheses
blank sigma equals negative fraction numerator 1 over denominator 4 end fraction
General
physics-

Four wires of the same material are stretched by the same load. Which one of them will elongate most if their dimensions are as follows

increment L equals fraction numerator F L over denominator A Y end fraction
Because, wires of the same material are stretched by the same load. So, F and Y will be constant.
therefore blank increment L proportional to fraction numerator L over denominator pi r to the power of 2 end exponent end fraction
increment L subscript 1 end subscript equals fraction numerator 100 over denominator pi cross times left parenthesis 1 cross times 10 to the power of negative 3 end exponent right parenthesis to the power of 2 end exponent end fraction equals fraction numerator 100 over denominator pi cross times 10 to the power of negative 6 end exponent end fraction
equals fraction numerator 100 over denominator pi end fraction cross times 10 to the power of negative 6 end exponent
therefore increment L subscript 2 end subscript equals fraction numerator 200 over denominator pi cross times left parenthesis 3 cross times 10 to the power of negative 3 end exponent right parenthesis to the power of 2 end exponent end fraction equals fraction numerator 200 over denominator pi cross times 9 cross times 10 to the power of negative 6 end exponent end fraction
equals fraction numerator 22.2 over denominator pi end fraction cross times 10 to the power of 6 end exponent
therefore increment L subscript 3 end subscript equals fraction numerator 300 over denominator pi cross times left parenthesis 3 cross times 10 to the power of negative 3 end exponent right parenthesis to the power of 2 end exponent end fraction equals fraction numerator 100 over denominator pi cross times 9 cross times 10 to the power of negative 6 end exponent end fraction
equals fraction numerator 33.3 over denominator pi end fraction cross times 10 to the power of 6 end exponent
therefore increment L subscript 4 end subscript equals fraction numerator 400 over denominator pi cross times left parenthesis 4 cross times 10 to the power of negative 3 end exponent right parenthesis to the power of 2 end exponent end fraction equals fraction numerator 400 over denominator pi cross times 16 cross times 10 to the power of negative 6 end exponent end fraction
equals fraction numerator 25 over denominator pi end fraction cross times 10 to the power of 6 end exponent
We can see that, L=100 cm and r = 1mm will elongate most.

Four wires of the same material are stretched by the same load. Which one of them will elongate most if their dimensions are as follows

physics-General
increment L equals fraction numerator F L over denominator A Y end fraction
Because, wires of the same material are stretched by the same load. So, F and Y will be constant.
therefore blank increment L proportional to fraction numerator L over denominator pi r to the power of 2 end exponent end fraction
increment L subscript 1 end subscript equals fraction numerator 100 over denominator pi cross times left parenthesis 1 cross times 10 to the power of negative 3 end exponent right parenthesis to the power of 2 end exponent end fraction equals fraction numerator 100 over denominator pi cross times 10 to the power of negative 6 end exponent end fraction
equals fraction numerator 100 over denominator pi end fraction cross times 10 to the power of negative 6 end exponent
therefore increment L subscript 2 end subscript equals fraction numerator 200 over denominator pi cross times left parenthesis 3 cross times 10 to the power of negative 3 end exponent right parenthesis to the power of 2 end exponent end fraction equals fraction numerator 200 over denominator pi cross times 9 cross times 10 to the power of negative 6 end exponent end fraction
equals fraction numerator 22.2 over denominator pi end fraction cross times 10 to the power of 6 end exponent
therefore increment L subscript 3 end subscript equals fraction numerator 300 over denominator pi cross times left parenthesis 3 cross times 10 to the power of negative 3 end exponent right parenthesis to the power of 2 end exponent end fraction equals fraction numerator 100 over denominator pi cross times 9 cross times 10 to the power of negative 6 end exponent end fraction
equals fraction numerator 33.3 over denominator pi end fraction cross times 10 to the power of 6 end exponent
therefore increment L subscript 4 end subscript equals fraction numerator 400 over denominator pi cross times left parenthesis 4 cross times 10 to the power of negative 3 end exponent right parenthesis to the power of 2 end exponent end fraction equals fraction numerator 400 over denominator pi cross times 16 cross times 10 to the power of negative 6 end exponent end fraction
equals fraction numerator 25 over denominator pi end fraction cross times 10 to the power of 6 end exponent
We can see that, L=100 cm and r = 1mm will elongate most.
General
physics-

A current i is flowing through the loop. The direction of the current and the shape of the loop are as shown in the figure.
The magnetic field at the centre of the loop is fraction numerator mu subscript 0 end subscript i over denominator R end fraction times.
left parenthesis M A equals R comma blank M B equals 2 R comma blank angle D M A equals 90 degree

(i) Magnetic field at the centre due to the curved portion D A equals fraction numerator mu subscript 0 end subscript i over denominator 4 pi R end fraction open parentheses fraction numerator 3 pi over denominator 2 end fraction close parentheses
According to right hand screw rule, the magnetic field will be into the plane of paper.

(ii) Magnetic field at M due to A B is zero.
(iii) Magnetic field at the centre due to the curved portion B C is fraction numerator mu subscript 0 end subscript i over denominator 4 pi 2 R end fraction open parentheses fraction numerator pi over denominator 2 end fraction close parentheses. According to
right hand screw rule, the magnetic field will be into the plane of paper.
(iv) Magnetic field at M due to D C is zero.
Hence, the resultant magnetic field at M
equals fraction numerator 3 mu subscript 0 end subscript i over denominator 8 R end fraction plus 0 plus fraction numerator mu subscript 0 end subscript i over denominator 16 R end fraction plus 0 equals fraction numerator 7 mu subscript 0 end subscript i over denominator 16 R end fraction

A current i is flowing through the loop. The direction of the current and the shape of the loop are as shown in the figure.
The magnetic field at the centre of the loop is fraction numerator mu subscript 0 end subscript i over denominator R end fraction times.
left parenthesis M A equals R comma blank M B equals 2 R comma blank angle D M A equals 90 degree

physics-General
(i) Magnetic field at the centre due to the curved portion D A equals fraction numerator mu subscript 0 end subscript i over denominator 4 pi R end fraction open parentheses fraction numerator 3 pi over denominator 2 end fraction close parentheses
According to right hand screw rule, the magnetic field will be into the plane of paper.

(ii) Magnetic field at M due to A B is zero.
(iii) Magnetic field at the centre due to the curved portion B C is fraction numerator mu subscript 0 end subscript i over denominator 4 pi 2 R end fraction open parentheses fraction numerator pi over denominator 2 end fraction close parentheses. According to
right hand screw rule, the magnetic field will be into the plane of paper.
(iv) Magnetic field at M due to D C is zero.
Hence, the resultant magnetic field at M
equals fraction numerator 3 mu subscript 0 end subscript i over denominator 8 R end fraction plus 0 plus fraction numerator mu subscript 0 end subscript i over denominator 16 R end fraction plus 0 equals fraction numerator 7 mu subscript 0 end subscript i over denominator 16 R end fraction
General
physics-

A part of a long wire carrying a current i is bent into a circle of radius r as shown in figure. The net magnetic field at the centre O of the circular loop is

The magnitude of the magnetic field at point O due to straight part of wire is
B subscript 1 end subscript equals fraction numerator mu subscript 0 end subscript i over denominator 2 pi r end fraction
B subscript 1 end subscript is perpendicular to the plane of the page, directed upwards (right hand plam rule 1).
The field at the centre O due to the current loop of radius r is
B subscript 2 end subscript equals fraction numerator mu subscript 0 end subscript i over denominator 2 r end fraction

B subscript 2 end subscript is also perpendicular to the page, directed upwards (right hand screw rule).
B subscript 1 end subscript plus B subscript 2 end subscript equals fraction numerator mu subscript 0 end subscript i over denominator 2 r end fraction open parentheses fraction numerator 1 over denominator pi end fraction plus 1 close parentheses
equals fraction numerator mu subscript 0 end subscript i over denominator 2 pi r end fraction open parentheses pi plus 1 close parentheses

A part of a long wire carrying a current i is bent into a circle of radius r as shown in figure. The net magnetic field at the centre O of the circular loop is

physics-General
The magnitude of the magnetic field at point O due to straight part of wire is
B subscript 1 end subscript equals fraction numerator mu subscript 0 end subscript i over denominator 2 pi r end fraction
B subscript 1 end subscript is perpendicular to the plane of the page, directed upwards (right hand plam rule 1).
The field at the centre O due to the current loop of radius r is
B subscript 2 end subscript equals fraction numerator mu subscript 0 end subscript i over denominator 2 r end fraction

B subscript 2 end subscript is also perpendicular to the page, directed upwards (right hand screw rule).
B subscript 1 end subscript plus B subscript 2 end subscript equals fraction numerator mu subscript 0 end subscript i over denominator 2 r end fraction open parentheses fraction numerator 1 over denominator pi end fraction plus 1 close parentheses
equals fraction numerator mu subscript 0 end subscript i over denominator 2 pi r end fraction open parentheses pi plus 1 close parentheses
General
physics-

A wire shown in figure carries a current of 40 A. If r equals 3.14 blank c m, the magnetic field at point P will be

Let the given circular A B C part of wire subtends an angle theta at its centre. Then, magnetic field due to this circular part is

B to the power of ´ ´ end exponent equals B subscript c end subscript cross times fraction numerator theta over denominator 2 pi end fraction equals fraction numerator mu subscript 0 end subscript over denominator 4 pi end fraction cross times fraction numerator 2 pi i over denominator e end fraction cross times fraction numerator theta over denominator 2 pi end fraction
rightwards double arrow blank B to the power of ´ ´ end exponent equals blank fraction numerator mu subscript 0 end subscript over denominator 4 pi end fraction blank. fraction numerator i over denominator r end fraction theta
Given, i equals 40 blank A comma blank r equals 3.14 blank c m equals 3.14 cross times 10 to the power of negative 2 end exponent blank m
theta equals 360 degree minus 90 degree equals 270 degree equals fraction numerator 3 pi over denominator 2 end fraction blank r a d.
therefore B to the power of ´ ´ end exponent equals fraction numerator 10 to the power of negative 7 end exponent cross times 40 over denominator 3.14 cross times 10 to the power of negative 2 end exponent end fraction cross times fraction numerator 3 pi over denominator 2 end fraction
B to the power of ´ ´ end exponent equals 6 cross times 10 to the power of negative 4 end exponent T

A wire shown in figure carries a current of 40 A. If r equals 3.14 blank c m, the magnetic field at point P will be

physics-General
Let the given circular A B C part of wire subtends an angle theta at its centre. Then, magnetic field due to this circular part is

B to the power of ´ ´ end exponent equals B subscript c end subscript cross times fraction numerator theta over denominator 2 pi end fraction equals fraction numerator mu subscript 0 end subscript over denominator 4 pi end fraction cross times fraction numerator 2 pi i over denominator e end fraction cross times fraction numerator theta over denominator 2 pi end fraction
rightwards double arrow blank B to the power of ´ ´ end exponent equals blank fraction numerator mu subscript 0 end subscript over denominator 4 pi end fraction blank. fraction numerator i over denominator r end fraction theta
Given, i equals 40 blank A comma blank r equals 3.14 blank c m equals 3.14 cross times 10 to the power of negative 2 end exponent blank m
theta equals 360 degree minus 90 degree equals 270 degree equals fraction numerator 3 pi over denominator 2 end fraction blank r a d.
therefore B to the power of ´ ´ end exponent equals fraction numerator 10 to the power of negative 7 end exponent cross times 40 over denominator 3.14 cross times 10 to the power of negative 2 end exponent end fraction cross times fraction numerator 3 pi over denominator 2 end fraction
B to the power of ´ ´ end exponent equals 6 cross times 10 to the power of negative 4 end exponent T
General
physics

A car covers one third part of its straight path with speed V1 and the rest with speed V2 . What is its average speed ?

A car covers one third part of its straight path with speed V1 and the rest with speed V2 . What is its average speed ?

physicsGeneral
General
physics

A bus travels between two points A and B. V1 and V2 are it average speed and average velocity then

A bus travels between two points A and B. V1 and V2 are it average speed and average velocity then

physicsGeneral
General
Maths-

If f colon R not stretchy rightwards arrow R is defined by f left parenthesis x right parenthesis equals fraction numerator x squared minus 4 over denominator x squared plus 1 end fraction text  then  end text f left parenthesis x right parenthesis text  is  end text

f colon R not stretchy rightwards arrow R comma space f left parenthesis x right parenthesis equals fraction numerator x squared minus 4 over denominator x squared plus 1 end fraction
t h e n comma space f left parenthesis x subscript 1 right parenthesis equals f left parenthesis x subscript 2 right parenthesis rightwards double arrow fraction numerator x subscript 1 squared minus 4 over denominator x subscript 1 squared plus 1 end fraction equals fraction numerator x subscript 2 squared minus 4 over denominator x subscript 2 squared plus 1 end fraction
rightwards double arrow x subscript 1 squared x subscript 2 squared minus 4 x subscript 2 squared plus x subscript 1 squared minus 4 equals x subscript 1 squared x subscript 2 squared minus 4 x subscript 1 squared plus x subscript 2 squared minus 4
rightwards double arrow 5 x subscript 1 squared space minus 5 x subscript 2 squared equals 0
rightwards double arrow left parenthesis x subscript 1 minus x subscript 2 right parenthesis plus left parenthesis x subscript 1 plus x subscript 2 right parenthesis equals 0
rightwards double arrow x subscript 1 equals x subscript 2 space o r space x subscript 1 equals space minus x subscript 2
s o comma space f space i s space n o t space o n e minus o n e.

l e t space y element of R comma
a n d space y equals fraction numerator x squared minus 4 over denominator x squared plus 1 end fraction rightwards double arrow x squared y plus y equals x squared minus 4 rightwards double arrow 4 plus y equals x squared minus x squared y rightwards double arrow x squared equals fraction numerator 4 plus y over denominator 1 minus y end fraction rightwards double arrow x equals square root of fraction numerator 4 plus y over denominator 1 minus y end fraction end root
f open parentheses square root of fraction numerator 4 plus y over denominator 1 minus y end fraction end root close parentheses equals fraction numerator open parentheses square root of fraction numerator 4 plus y over denominator 1 minus y end fraction end root close parentheses squared minus 4 over denominator open parentheses square root of fraction numerator 4 plus y over denominator 1 minus y end fraction end root close parentheses squared plus 1 end fraction equals fraction numerator 4 plus y minus 4 plus 4 y over denominator 4 plus y plus 1 minus y end fraction equals fraction numerator 5 y over denominator 5 end fraction equals y
S o comma space f space i s space o n t o.

If f colon R not stretchy rightwards arrow R is defined by f left parenthesis x right parenthesis equals fraction numerator x squared minus 4 over denominator x squared plus 1 end fraction text  then  end text f left parenthesis x right parenthesis text  is  end text

Maths-General
f colon R not stretchy rightwards arrow R comma space f left parenthesis x right parenthesis equals fraction numerator x squared minus 4 over denominator x squared plus 1 end fraction
t h e n comma space f left parenthesis x subscript 1 right parenthesis equals f left parenthesis x subscript 2 right parenthesis rightwards double arrow fraction numerator x subscript 1 squared minus 4 over denominator x subscript 1 squared plus 1 end fraction equals fraction numerator x subscript 2 squared minus 4 over denominator x subscript 2 squared plus 1 end fraction
rightwards double arrow x subscript 1 squared x subscript 2 squared minus 4 x subscript 2 squared plus x subscript 1 squared minus 4 equals x subscript 1 squared x subscript 2 squared minus 4 x subscript 1 squared plus x subscript 2 squared minus 4
rightwards double arrow 5 x subscript 1 squared space minus 5 x subscript 2 squared equals 0
rightwards double arrow left parenthesis x subscript 1 minus x subscript 2 right parenthesis plus left parenthesis x subscript 1 plus x subscript 2 right parenthesis equals 0
rightwards double arrow x subscript 1 equals x subscript 2 space o r space x subscript 1 equals space minus x subscript 2
s o comma space f space i s space n o t space o n e minus o n e.

l e t space y element of R comma
a n d space y equals fraction numerator x squared minus 4 over denominator x squared plus 1 end fraction rightwards double arrow x squared y plus y equals x squared minus 4 rightwards double arrow 4 plus y equals x squared minus x squared y rightwards double arrow x squared equals fraction numerator 4 plus y over denominator 1 minus y end fraction rightwards double arrow x equals square root of fraction numerator 4 plus y over denominator 1 minus y end fraction end root
f open parentheses square root of fraction numerator 4 plus y over denominator 1 minus y end fraction end root close parentheses equals fraction numerator open parentheses square root of fraction numerator 4 plus y over denominator 1 minus y end fraction end root close parentheses squared minus 4 over denominator open parentheses square root of fraction numerator 4 plus y over denominator 1 minus y end fraction end root close parentheses squared plus 1 end fraction equals fraction numerator 4 plus y minus 4 plus 4 y over denominator 4 plus y plus 1 minus y end fraction equals fraction numerator 5 y over denominator 5 end fraction equals y
S o comma space f space i s space o n t o.
General
physics

A car moving over a straight path covers a distance x with constant speed 10ms-1 and then the same distance with constant speed of V1.If average speed of the car is , 16 ms-1 then V2=...

A car moving over a straight path covers a distance x with constant speed 10ms-1 and then the same distance with constant speed of V1.If average speed of the car is , 16 ms-1 then V2=...

physicsGeneral
General
physics-

A copper wire of negligible mass, 1 m length and cross-sectional area 10 to the power of negative 6 end exponent is kept on a smooth horizontal table with one end fixed. A ball of mass 1 kg is attached to the other end. The wire and the ball are rotated with an angular velocity 20 rads to the power of negative 1 end exponent. blankIf the elongation in the wire is 10 to the power of negative 3 end exponent m, then the Young’s modulus is

Y equals fraction numerator F l over denominator A increment l end fraction equals blank fraction numerator open parentheses m l omega to the power of 2 end exponent close parentheses l over denominator A blank increment l end fraction o r blank Y equals fraction numerator m l to the power of 2 end exponent omega to the power of 2 end exponent over denominator A increment l end fraction
Or Y equals blank fraction numerator 1 cross times 1 cross times 1 cross times 20 cross times 20 over denominator 10 to the power of negative 6 end exponent cross times 10 to the power of negative 3 end exponent end fraction equals 4 blank cross times 10 to the power of 11 end exponent N m to the power of negative 2 end exponent

A copper wire of negligible mass, 1 m length and cross-sectional area 10 to the power of negative 6 end exponent is kept on a smooth horizontal table with one end fixed. A ball of mass 1 kg is attached to the other end. The wire and the ball are rotated with an angular velocity 20 rads to the power of negative 1 end exponent. blankIf the elongation in the wire is 10 to the power of negative 3 end exponent m, then the Young’s modulus is

physics-General
Y equals fraction numerator F l over denominator A increment l end fraction equals blank fraction numerator open parentheses m l omega to the power of 2 end exponent close parentheses l over denominator A blank increment l end fraction o r blank Y equals fraction numerator m l to the power of 2 end exponent omega to the power of 2 end exponent over denominator A increment l end fraction
Or Y equals blank fraction numerator 1 cross times 1 cross times 1 cross times 20 cross times 20 over denominator 10 to the power of negative 6 end exponent cross times 10 to the power of negative 3 end exponent end fraction equals 4 blank cross times 10 to the power of 11 end exponent N m to the power of negative 2 end exponent
General
physics-

Two wires of equal cross-section but one made of steel and the other of copper are joined end to end. When the combination is kept under tension, the elongations in the two wires are found to be equal. What is the ratio of the lengths of the two wires? (Given : steel = 2 cross times 10 to the power of 11 end exponent N m to the power of negative 2 end exponent right parenthesis

Y equals fraction numerator s t r e s s over denominator s t r a i n end fraction blank o r blank s t r a i n equals fraction numerator s t r e s s over denominator Y end fraction o r fraction numerator increment L over denominator L end fraction equals fraction numerator s t r e s s over denominator Y end fraction
Since, cross-sections are equal and same tension exists in both the wires, therefore, the stresses developed are equal.
Also, increment L is given to be the same for both the wires.
therefore blank L blank proportional to Y
therefore blank fraction numerator L subscript s end subscript over denominator L subscript C u end subscript end fraction equals blank fraction numerator 2 blank cross times 10 to the power of 11 end exponent over denominator 1.1 blank cross times 10 to the power of 11 end exponent end fraction equals fraction numerator 20 over denominator 11 end fraction

Two wires of equal cross-section but one made of steel and the other of copper are joined end to end. When the combination is kept under tension, the elongations in the two wires are found to be equal. What is the ratio of the lengths of the two wires? (Given : steel = 2 cross times 10 to the power of 11 end exponent N m to the power of negative 2 end exponent right parenthesis

physics-General
Y equals fraction numerator s t r e s s over denominator s t r a i n end fraction blank o r blank s t r a i n equals fraction numerator s t r e s s over denominator Y end fraction o r fraction numerator increment L over denominator L end fraction equals fraction numerator s t r e s s over denominator Y end fraction
Since, cross-sections are equal and same tension exists in both the wires, therefore, the stresses developed are equal.
Also, increment L is given to be the same for both the wires.
therefore blank L blank proportional to Y
therefore blank fraction numerator L subscript s end subscript over denominator L subscript C u end subscript end fraction equals blank fraction numerator 2 blank cross times 10 to the power of 11 end exponent over denominator 1.1 blank cross times 10 to the power of 11 end exponent end fraction equals fraction numerator 20 over denominator 11 end fraction
General
physics-

The relation between gamma comma eta and K for a elastic material is

The relation between gamma comma eta and K for a elastic material is

physics-General
General
Maths-

If f colon R not stretchy rightwards arrow R is defined by f left parenthesis x right parenthesis equals fraction numerator x squared minus 4 over denominator x squared plus 1 end fraction then f left parenthesis x right parenthesis text  is  end text

If f colon R not stretchy rightwards arrow R is defined by f left parenthesis x right parenthesis equals fraction numerator x squared minus 4 over denominator x squared plus 1 end fraction then f left parenthesis x right parenthesis text  is  end text

Maths-General
General
physics-

Two long parallel conductors carry currents in opposite directions as shown. One conductor carries a current of 10 A and the distance between the wires is d=10cm  Current I is adjusted, so that the magnetic field at P is zero. P is at a distance of 5 cm to the right of the 10 A current. Value of I is

From Biot-Savart’s law the magnetic field open parentheses B close parentheses due to a conductor carrying current I comma at a distance r subscript 1 end subscript is
B subscript 1 end subscript equals fraction numerator mu subscript 0 end subscript I subscript 1 end subscript over denominator 2 pi r subscript 1 end subscript end fraction
Magnetic field at P due to current in second conductor is
B subscript 2 end subscript equals fraction numerator mu subscript 0 end subscript I subscript 2 end subscript over denominator 2 pi open parentheses r subscript 1 end subscript plus d close parentheses end fraction
From Fleming’s right hands rule the fields at P are directed opposite.
therefore R e s u l t a n t s comma blank f i e l d blank B subscript 1 end subscript equals B subscript 2 end subscript
therefore blank fraction numerator mu subscript 0 end subscript I subscript 1 end subscript over denominator 2 pi r subscript 1 end subscript end fraction equals blank fraction numerator mu subscript 0 end subscript I subscript 2 end subscript over denominator 2 pi open parentheses r subscript 1 end subscript plus d close parentheses end fraction
Given, I subscript 1 end subscript equals 10 blank A comma blank r subscript 1 end subscript equals 5 comma blank r subscript 1 end subscript plus d equals 5 plus 10 equals 15 blank c m
therefore blank I subscript 2 end subscript equals fraction numerator I subscript 1 end subscript over denominator r subscript 1 end subscript end fraction blank cross times open parentheses r subscript 1 end subscript plus d close parentheses
I subscript 2 end subscript equals fraction numerator 10 over denominator 5 end fraction cross times 15 equals 30 blank A

Two long parallel conductors carry currents in opposite directions as shown. One conductor carries a current of 10 A and the distance between the wires is d=10cm  Current I is adjusted, so that the magnetic field at P is zero. P is at a distance of 5 cm to the right of the 10 A current. Value of I is

physics-General
From Biot-Savart’s law the magnetic field open parentheses B close parentheses due to a conductor carrying current I comma at a distance r subscript 1 end subscript is
B subscript 1 end subscript equals fraction numerator mu subscript 0 end subscript I subscript 1 end subscript over denominator 2 pi r subscript 1 end subscript end fraction
Magnetic field at P due to current in second conductor is
B subscript 2 end subscript equals fraction numerator mu subscript 0 end subscript I subscript 2 end subscript over denominator 2 pi open parentheses r subscript 1 end subscript plus d close parentheses end fraction
From Fleming’s right hands rule the fields at P are directed opposite.
therefore R e s u l t a n t s comma blank f i e l d blank B subscript 1 end subscript equals B subscript 2 end subscript
therefore blank fraction numerator mu subscript 0 end subscript I subscript 1 end subscript over denominator 2 pi r subscript 1 end subscript end fraction equals blank fraction numerator mu subscript 0 end subscript I subscript 2 end subscript over denominator 2 pi open parentheses r subscript 1 end subscript plus d close parentheses end fraction
Given, I subscript 1 end subscript equals 10 blank A comma blank r subscript 1 end subscript equals 5 comma blank r subscript 1 end subscript plus d equals 5 plus 10 equals 15 blank c m
therefore blank I subscript 2 end subscript equals fraction numerator I subscript 1 end subscript over denominator r subscript 1 end subscript end fraction blank cross times open parentheses r subscript 1 end subscript plus d close parentheses
I subscript 2 end subscript equals fraction numerator 10 over denominator 5 end fraction cross times 15 equals 30 blank A
General
physics-

Two long straight wires are set parallel to each other. Each carries a current i in the opposite direction and the separation between them is 2R. The intensity of the magnetic field midway between them is

Magnetic field at mid-point due to wire A B
B subscript 1 end subscript equals fraction numerator mu subscript 0 end subscript i over denominator 2 pi R end fraction
Magnetic field at mid-point due to wire C D
B subscript 2 end subscript equals fraction numerator mu subscript 0 end subscript i over denominator 2 pi R end fraction
Resultant of Magnetic field B equals B subscript 1 end subscript plus B subscript 2 end subscript
equals fraction numerator mu subscript 0 end subscript i over denominator 2 pi R end fraction plus fraction numerator mu subscript 0 end subscript i over denominator 2 pi R end fraction
B equals fraction numerator mu subscript 0 end subscript i over denominator pi R end fraction

Two long straight wires are set parallel to each other. Each carries a current i in the opposite direction and the separation between them is 2R. The intensity of the magnetic field midway between them is

physics-General
Magnetic field at mid-point due to wire A B
B subscript 1 end subscript equals fraction numerator mu subscript 0 end subscript i over denominator 2 pi R end fraction
Magnetic field at mid-point due to wire C D
B subscript 2 end subscript equals fraction numerator mu subscript 0 end subscript i over denominator 2 pi R end fraction
Resultant of Magnetic field B equals B subscript 1 end subscript plus B subscript 2 end subscript
equals fraction numerator mu subscript 0 end subscript i over denominator 2 pi R end fraction plus fraction numerator mu subscript 0 end subscript i over denominator 2 pi R end fraction
B equals fraction numerator mu subscript 0 end subscript i over denominator pi R end fraction