Magnetic Field Due To A Current

An electric current flows in a wire from north to south. So the direction of the magnetic field associated with it can be predicted by Maxwell’s Right hand Cork-Screw rule.

So we can conclude that,

(a) The magnetic field to the east of the wire will be directed vertically upwards.

(b) The magnetic field to the west of the wire will be directed vertically downwards.

(c) The magnetic field vertically above the wire will be directed towards the west.

(d) The magnetic field vertically below the wire will be directed towards the east.

Magnetic field at a point d distance apart from a long straight current carrying wire is given by from Ampere’s circuital law. Where μ₀ is the magnetic permeability of free space, i denotes the current through the wire and d denotes the distance from the wire, of the point where magnetic field is to be calculated.

Now we know that, speed of light, (ϵ₀ denotes the electric permittivity of free space)

c² = μ₀ ϵ₀

Thus we get magnetic field to be

We can predict the direction of the magnetic field by Maxwell’s Right hand Cork-Screw rule. So we put the thumb of our right hands pointing towards the direction of current and now the direction the other fingers curl toward will be direction of the magnetic field.

So, we get the field at the center is moving away from me if the current is going in the clockwise direction.

The contribution to the magnetic field B by currents outside the Amperian loop turns out to be zero because if we draw a straight line on any fixed point on the loop from the current vector(the direction vector of current), let line A and draw another line from the current vector to a movable point on the loop, let line B, and move that point in one direction along the loop until the full loop is traversed then we can see the net positive angle travelled by line B with respect to line A is equal to the net negative value travelled by the same. So the total angle travelled is zero. The line integral has a finite value for a current that is within the loop as it traverses an angle of 2ϖ. So, we can say that Ampere’s law takes into account every current be it inside the Amperian loop/curve or outside but as for outside currents, the integral is zero, so only the enclosed current is taken into account for calculation but actually takes into consideration the contribution of all the currents.

we consider the circle to be the Amperian loop. Point B is the fixed point C is the moving point. At some point of time C coincides with C’ where ∠BAC = -∠BAC’. We will get a point like this for every arbitrary point on the circle, so total angle for all the circle will be zero as stated before.

Magnetic field on the axis of a circular current carrying loop is given by,

where a denotes the radius of the loop and x is the distance of the point on the axis at which the magnetic field is to be calculated. i is the current flowing in the loop. μ₀ is the magnetic permeability of free space.

And B = μ₀ni (where μ₀ is the magnetic permeability of free space, n is the number of turns of the wire per unit length of the solenoid, and i is current in the wire) holds true for a very long solenoid where at the terminal points x >> a. We can say that if we keep adding more loops at the ends, the total number of loops increase but with that length also increases so the turns per unit length n is unvaried. We can also think of it this way,

The value of B for loops added to the ends of the solenoid will have dropped to almost zero as for the condition x >> a,

This can be shown mathematically by taking the limit of B for a circular loop. So, let us take x to be tending to infinity and a to have a finite value.

(Dividing both sides by x³)

So if we add more loops at the ends of a very long solenoid, the magnetic field due to each one of them will be zero at the centre of the solenoid thus won’t have considerable effect.

Ampere’s law holds true for any closed loop of magnetic field containing constant electric current. The left hand side of Ampere’s circuital law

Where

μ₀ is the magnetic permeability of free space,

i is the current enclosed by the loop and B is the magnetic field

The above equation deals with a closed line integral along the path of the magnetic field. The line integral will not have a finite value if the loop is not closed. So, Ampere’s law will not hold for a loop that does not enclose the wire but it will hold for any shape of the loop even if it’s not circular.

No, there won’t be any magnetic force acting on the wire in this case because, if the ring starts rotating, it will create a magnetic field which is parallel or antiparallel to the direction of current or the length vector of the wire at the center of the ring. i.e. angle between the is zero or ϖ respectively. According to F = ILBsinϴ

for ϴ = 0 or ϖ,

F= 0.

Here I is the current, L is the length of the wire and B is the magnitude of the magnetic field.

So, there will be no force on the wire if the ring starts rotating about the wire.

Using the relation for force acting on a conductor

Where

I is the current in the wire,

L is the length vector of the wire

B is the external electric field.

Thus, if the wires are kept perpendicular, the angle between them will be 90 and thus the angle between the movable wire and the magnetic field due to the fixed wire is 0( from Maxwell’s Cork-Screw rule). Thus the force is the minimum and it is in an unstable equilibrium. Thus the will rotate and make itself parallel to the fixed wire and attain the maximum attractive force and thus a stable equilibrium.

The magnetic field is not as strong as electric field and the magnitude of magnetic field due to a moving charge only equals that of its electric field only when the velocity is equal to the speed of light. So for two protons moving in the same direction, Force due to electric field repulsion will exceed that due to magnetic field attraction. So the proton beams moving in the same direction repel each other.

But two parallel current carrying wire attract due to magnetic attraction as the electric field is inside the conductor and is shielded by it and can’t be felt outside. The magnetic attraction can be further reinforced by Fleming’s left rule by considering the magnetic field produced by one wire and electric field in the other wire in the direction of current.

We can always find the magnetic field due to a long, straight current carrying wire along any loop which only contains that wire by using Ampere’s circuital law. The law only takes into account the current bound in the loop irrespective of the source. If we take the whole circuit along with the battery within that Amperian loop, then the total current contained in that loop will be zero so the resultant magnetic field will be zero too even if the wire itself produces a non zero magnetic field.

If we choose an Amperian loop containing both the wires, then the magnetic field along that loop will turn out to be zero as the wires carry currents in the opposite direction and the total current turns out to be lesser than the maximum of that of the two wires. And if the magnitudes of currents are the same then the total current in that loop turns out to be zero. If they are kept very close together, this will hold for even very small Amperian loops. Thus twisting two wires, which carry currents in the opposite directions, minimizes the magnetic field produced and can even cancel it out.

denotes current going perpendicularly inside the screen.

denotes current coming perpendicularly out of the screen.

If this both currents are equal in magnitude then the total current enclosed by the Amperian Loop C will be zero.

The displacement of the wire due to magnetic forces happens in the direction of the distance vector between the two wires. The magnetic field due to the wires is perpendicular to the wire and the distance vector between the wires so work done is by the field is zero as usual. As the wires move towards each other there is a change in magnetic field and according to Lenz law there will be an induced current in the wires which will be directed opposite to the current initially carried by them. As these electrons cannot come out of the wire, they hit the edges of the wire starts moving and gains kinetic energy.

With the right hand cork screw we can get the direction of the magnetic field due to the vertical current carrying wire. And we find it to be directed downwards. Now applying Fleming’s Left hand rule, we can find the direction of force as we put the direction of the electron beam as the opposite direction of current. We can find that force is pointed upwards. So the electron beam will be deflected upwards.

There won’t be any magnetic force acting on the wire in this case because, according the Cork-Screw rule, the magnetic field due to the straight wire will be parallel or anti parallel to the direction of current in the loop, thus creating an angle of 0 or ϖ with the direction of current.

According to F = ILBsinϴ

Where

I is the current,

L is the length of the wire

B is the magnitude of the magnetic field.

for ϴ = 0 or ϖ, F= 0.

We can consider the beams to be current carrying wires. The proton beam is going from north to south so the current for it will be directed from north to south. And the electron beam is moving from South to north so the current will be directed opposite to the direction of the beam, i.e. from north to south. So we get the case of two current carrying wires carrying current in the same direction and they attract each other. So the electron beam will be deflected towards the proton beam.

If the current is directed towards north at the topmost point of that circular loop that means, from point A’s perspective, current is in clockwise direction and will cause a magnetic field which is moving inwards i.e. to the west at point A and subsequently to the west at point B.

The closest side of the loop to the straight wire carries a current in the same direction as the straight wire. So there will be an attractive force. And for the opposite side of the loop the current is moving in the opposite direction to that of the straight wire, so there will be a repulsive force. As the closest side shows attractive force and the farthest side shows repulsive force, the attractive force will be dominant. To explain it mathematically, we introduce the expression for force per unit length on each of the wires.

Where

F denote the force per unit length,

I₁, I₂ denote the currents in the two wires,

d is the distance between them

μ₀ is the magnetic permeability of free space.

The current in the loop is of same magnitude everywhere. The parallel sides of the loop to the long straight wire will encounter all the force. As the side closer to the straight wire carries a current in the same direction, I₁I₂ will be positive thus it will encounter attractive force and the opposite side will encounter repulsive force as the, I₁I₂ is negative as the currents are in opposite directions. We can see that the force is inversely proportional to the distance. As the wire carrying current in the same direction is closer than the wire carrying current in the opposite direction, the attractive force will be greater than the repulsive force. The other two sides will experience forces of equal amplitude but opposite directions and thus will cancel each other out. Thus the loop will move towards the wire.

If a charged particle is moving along a magnetic field line, that means it is moving either parallel or anti parallel to the direction of magnetic field. i.e. the direction of velocity of the particle is either making an angle of 0 or ϖ. According to F = qvBsinϴ (where q is the charge of the particle, v is the velocity of it and B is the magnitude of the external Magnetic field and ϴ is the angle between the direction of magnetic field and the velocity) for ϴ = 0 or ϖ, F= 0.

Under any circumstances, a charge particle will produce an electric field. And if we imagine that the moving charge particle is analogous to an electric current then it is clear to us that it, most certainly, will produce a magnetic field as well.

If a charged particle is projected with velocity v on a plane perpendicular to a uniform magnetic field B, then the particle will travel in a circle. Whose radius is where m and q are the mass and the charge of the particle respectively.

So the area enclosed by the path will be A = ϖR² = = = kinetic energy

Thus, the area will be proportional to the kinetic energy of the particle.

let the charge of both of them be q and mass be m_x and m_y respectively. If they are driven through the same potential V then,

Thus, v_x and v_y (their respective velocities after leaving the potential) will be and

So from we get, (all the other terms are constant for both the particles)

So, according to the question

Thus,

The third wire is carrying a current 20A antiparallel to the previously set 20A current wire. So these wires will repel each other. As the 40A wire is carrying a current in the opposite direction to the previously set 20A wire, the newly set 20A wire will be having the same current direction to it. So they will attract. So the newly set wire in the midway will move towards the 40A wire.

The magnetic field caused by a long wire is given by where i is the current in the wire, d is the distance at which B is to be calculated, μ₀ is the magnetic permeability of free space. Let the distance between the two wires be 2d.

Then, when the currents are parallel, one’s magnetic field will negate the other’s because at the point midway between the two wires, the magnetic fields generated by them will be directed in the opposite directions and it can be verified by the right hand Cork-Screw rule.

So the total B will be

which is given to be 10μT and when the direction of i₂ is reversed then, is given to be 30μT. So, solving these, we get,

(Componendo- Dividendo rule)

We know, drift velocity of electron is

vd = i/(nAe)

Where

i is the current in the wire,

n is the electron number density of the wire a,

A is the area of cross section of the wire,

e is the charge of the electron.

We can see that the trolley is moving with drift velocity but current travels in a conductor with a speed comparable to that of light in free space. So the velocity of the trolley is very low compared to the velocity at which the current travels. So there will be no significant change in the value of B and it will be .

According to Biot-Savart law:
The magnetic field due to a current element is given as
Where,

i is the current,

is the length element vector,

is the vector joining the required point and the length element,

μ₀ is the permeability of vacuum and μ₀ = 4π × 10^-7 TmA^-1.
Also, by vector properties: A× B = -(B× A)
Thus,
Thus, options (A) and (B) are the correct options

(A)Dimension for electric field E is: F/q = [MLT^-3A^-1]
Force due to magnetic field is given as: F=qvB.
Where q is the charge, v is the velocity of charge and B is the magnetic field.
Hence

The dimensions of charge, q is [AT]

The dimension of the velocity, V is [LT-1]

Putting the value in the above formula, we get

Dimensions of B =

∴ Dimensions of B = [MT^-2A^-1]
Thus,

(B)
We know that Where

μ₀ is permeability of free space,

ϵ₀ is the permittivity of free space and

c is the speed of light.
Thus, y = c, c is the speed and unit of speed is m/s
Dimensions of y = [LT^-1]
(C)

Reactance,
Where

C is the capacitance

R is the resistance

l is the length of the wire.

In a CR circuit the product CR is the time constant and unit is seconds.
Thus, Dimensions of z = l/CR = m/s= [LT^-1]
Hence x, y and z have same dimensions.
Thus, options (A),(B) and (C) are the correct options.

When a current carrying wire is along z-axis, the magnetic field can be found using right hand thumb rule.

Here X represents the field is going in and O represents field is coming out.
First case, if two points in the XY plain are above the wire and at same distance from z-axis, then both will lie in the X region(field going in), thus have same direction of magnetic field and magnitudes of magnetic field would be same.
Second case , if the two points in XY plane are at same distance from z axis but one point is above the wire and one is below it, then the direction of the electric field would be opposite for both the points. One point having X direction (going in) and other having O direction (coming out).
First two cases satisfy options (B),(C) and (D).
Thus, options (B),(C) and (D) are the correct options.

According to Ampere’s Law:
Where,

dl is the current element,

B is the magnetic field,

μ₀ is the permeability of free space and

i is the current flowing.

Thus, at the cross-section the formula becomes,
2πR is the circumference of the wire and R is the radius. We get,
Now, at the axis of the wire. R=0, and so no area to integrate and hence zero current is enclosed. Thus magnitude of magnetic field is minimum at axis of the wire.
At the surface of the wire, R= some minimum value.
As R increases , magnitude of magnetic field decreases.
Hence, B will be maximum at the surface of the wire.
Thus, options (B) and (C) are correct options.

By Ampere’s Law ,
Here, dl is the current element, B is the magnetic field, μ₀ is the permeability of free space and i is the current flowing.
We know that , magnetic field due to a circular wire is given
R is the radius of the wire.
Consider a Hollow tube as the series of circular wires.
Inside the tube at a point, r<R
Current inside the hollow tube is zero: i_inside = 0

Hence , the magnetic field is constant inside the hollow tube.
Now, at the axis, R=0.

Therefore, zero current is enclosed.
Hence, B=0.
Thus, options (B) and (C) are the correct options.

According to Ampere’s Law:
Where,

B is the magnetic field,

dl is the current element,

μ₀ is the permeability of free space

i is the current flowing in the conductor

Magnetic field outside the cable: B = B_I+B_O
B_I and B_O are the magnetic field of the inner conductor and outer cable respectively.
Due to wire,
R is the radius of the wire.


Thus, option (A) is the correct option.

Magnetic field inside the inner conductor:

∮B. dl =μ_o I (I=0(inside the wire))

Magnetic field between the two conductors and inside the outer conductor.

∮B. dl =μ_o I

On integrating the length element over the complete loop, we get dl= 2π r

B.2π r=μ_o I

On solving the above equation for B, we get

Thus, option A and B is the correct option.

Using Ampere’s law, the electric field due to a current carrying conductor of cross section areas is:
magnetic field,
Where,

dl is the current element,

μ₀ is the permeability of free space,

i is the current flowing

R is the radius of the conductor.
At the axis,

R=0 and so no area to enclose current.

Hence, B=0 at the axis of the conductor.

Electric field in the vicinity of the conductor is zero as a minimum distance between the point and the conductor is required to have electric field at that point.
Thus, options (B) and (C) are correct options.

Given:

Formula used:
Lorentz force:
Where, is the magnetic force vector, q is the charge, is the velocity vector and is the magnetic field vector.

Magnetic field can be written as:
Unit of force is Newton: N
Charge: q = It
Unit of current : I = Amperes= A
Unit of time : t = seconds = s
Unit of charge : q = As
Unit of velocity : v = m/s
Thus, unit of Magnetic field: Therefore, unit of Magnetic field : B = N m^-1A^-1.
Now,
Ampere’s Law:
Where,

μ₀ is the permeability of free space,

I is the current flowing

r is the radius of the current carrying wire.
SI Unit of B = N mA^-1
SI Unit of I = A
SI Unit of r = m


Thus, unit of permeability constant μ₀ is N A^-2.
Hence proved that unit of Magnetic field is N m^-1A^-1 and unit of permeability constant μ₀ is N A^-2.

Given:
Current : I = 10 A

Point is located along x-axis at 1 m.
Formula used:
Using Ampere’s Law for a current carrying straight wire:
Here, B is the magnitude of magnetic field, μ₀ is the permeability of free space and μ₀= 4π × 10^-7 T mA^-1 and d is the distance between the current carrying wire and the required point.
Substituting the values,

∴B = 2×10^-6 T
Hence, magnetic field at a point (1m, 0, 0) is 2× 10^-6 T.

Given:
Diameter of the copper wire: d = 1.6mm = 1.6× 10^-3 m
Current through the wire: I = 20 A
Formula used:
By Ampere’s law for current carrying wire of cross section area,
Where,

B is the magnitude of magnetic field,

μ₀ is the permeability of free space and μ₀= 4π × 10^-7 T mA^-1 r is the radius of the wire.

Radius of the wire : r = d/2 = 8× 10^-3 m
Substituting we get,

∴ B = 5× 10^-3 T
Hence, maximum magnitude of magnetic field due to a current of 20 A is 5× 10^-3 T.

Given:
Current through the transmission wire: I = 100 A
Distance between wire and the road : d = 8 m
Formula used:
By Ampere’s Law for a current carrying wire is
Where,

B is the magnitude of magnetic field,

μ₀ is the permeability of free space

μ₀= 4π × 10^-7 T mA^-1

d is the distance between the current carrying wire and the required point.
Substituting we get,

B = 2.5× 10^-6 T
Hence, magnitude of magnetic field at a point on the road due to current carrying transmission wire is 2.5× 10^-6 T.

Given :
Current in the wire : i = 1.0 A
Magnitude of Horizontal magnetic field: B_o = 1.0 × 10^–5 T
Distance between points P,Q and the wire :
d = 2.0 cm = 0.02 m

Right hand rule is used to determine the direction of the magnetic field due to the current carrying wire. Cross determines the field going into the plane and dot determines the field coming out of the field.
B_w is the magnetic field due to wire.
Formula used:
By Ampere’s Law for a current carrying wire is
Where,

B is the magnitude of magnetic field,

μ₀ is the permeability of free space, μ₀= 4π × 10^-7 T mA^-1

d is the distance between the current carrying wire and the required point.

Magnetic field at point P due to wire,


Resultant magnetic field at P
B_P = B_o +B_w
∴ B_P = 1× 10^-5 + 1× 10^-5
∴ B_P = 2 × 10^-5 T
Magnetic field at point Q due to wire,

∴ B_w = - 1 × 10^-5 T
The negative sign is because the field at Q is opposite to the direction of horizontal field and at P.
Resultant magnetic field at P
B_Q = B_o +B_w
∴ B_Q = 1× 10^-5 - 1× 10^-5
∴ B_Q = 0
Hence, resultant magnetic field at P is 2 × 10^-5 T and at Q is 0.

Given:
Radius of the wire : r
Current through the wire : i
Uniform magnetic field pointing upward : B

Formula used:
Magnetic field due to a current carrying wire is

Magnitude of magnetic field due to wire,

Where, μ₀ is the permeability of free space and μ₀= 4π × 10^-7 T mA^-1.

r is the radius of the wire.
(a)
Maximum net magnetic field would be above the wire at P.
Considering direction of the current as shown in the diagram, right hand rule will give us direction of the magnetic field due to wire.
thus,
B_net = B+B_w
∴ B_net = B +
(b)
Similarly, the field at Q would be in opposite direction to the P.
Minimum magnitude of the magnetic field will be a Q
B_net = B-B_w
∴ B_net = B -
Hence, magnitude of magnetic field will be maximum at points above at wire and minimum at points below the wire only if the direction of the current is as shown in the figure.

Given:
Current in the wire : I = 30 A
External Uniform magnetic field: B_O = 4.0 ×10^–4 T
Distance between the point and the wire : d = 2 cm = 0.02 m

Here, B is the magnetic field due to wire. B_O is parallel to the current in the wire. Using Right hand rule we can see that direction of magnetic field at the desired point is going into the plane (cross). This direction due to wire is perpendicular to the External magnetic field.
Formula used:
By Ampere’s Law for a current carrying wire is
Here, B is the magnitude of magnetic field, μ₀ is the permeability of free space and μ₀= 4π × 10^-7 T mA^-1 and d is the distance between the current carrying wire and the required point.
Substituting we get,

∴ B = 3× 10^-4 T
As Magnetic field due to wire is perpendicular to the external magnetic field, the resultant net magnetic field would be:


∴ B_net = 5 × 10^-4 T
Hence, the magnitude of the resultant magnetic field at a point 2.0 cm away from the wire is 5 × 10^-4 T.

Given:
Current in the wire : I = 10 A
Magnitude of Horizontal magnetic field : B_O = 2.0 × 10^–3 T

For the resultant magnetic field to be zero, we need to have a field in opposite direction to that of the existing field.
Hence, magnetic field B due to wire in the direction from north to south as shown in the diagram..
Formula used:
By Ampere’s Law for a current carrying wire is
Where,

B is the magnitude of magnetic field,

μ₀ is the permeability of free space and μ₀= 4π × 10^-7 T mA^-1 d is the distance between the current carrying wire and the required point.
To get zero resultant,
B=B_O


∴ d = 0.001 m=1 mm
Hence, the point should be place at a distance of 0.001 m from the wire in west direction

Given:
Magnitude of current in both the wires: i = 10 A
Distance between 2 wires : d = 4cm = 0.04 m

P and Q are the two wires having opposite direction of currents. From the diagram, P has current into the plane and Q has current coming out of the plane.
By right hand rule, P will have field coming up at A₁ and going down at A₂ and A₃ and tangent to A₄ as shown by the red arrow.
Similarly, Q will have field going down as A_1,A₂ and A₃ aand tangent to A₄ as shown by blue arrow.
Distance between, A₁,A₂ and A₃ is 2 cm each.
Formula used:

By Ampere’s Law for a current carrying wire is
Here, B is the magnitude of magnetic field, μ₀ is the permeability of free space and μ₀= 4π × 10^-7 T mA^-1 and d is the distance between the current carrying wire and the required point.
At A₁ :
Net magnetic field due to P and Q would be subtraction of individual fields at A₁.
PA₁ :d₁ = 0.02 m
QA₁ : d₂ = 0.06m
B_net = B_P – B_Q
Where, B_P is the magnetic field due to wire P and B_Q is the magnetic field due to wire Q.


∴ B_net = 6.66 × 10^-5 T
Hence, magnetic field at A₁ is 6.66 × 10^-5 T.
At A₂
Effect of magnetic field at A₂ will be sum of magnetic field due to P and Q as both are in same direction as shown.
PA₂: d₁ = 0.01 m
QA₃: d₂ = 0.03 m
B_net = B_P + B_Q
Where, B_P is the magnetic field due to wire P and B_Q is the magnetic field due to wire Q.



∴ B_net = 2.66 × 10^-4 T
Hence , magnetic field at A₂ is 2.66 × 10^-4 T
At A₃
Effect of magnetic field at A₃ will be sum of magnetic field due to P and Q as both are in same direction as shown.
PA₃: d₁ = 0.02 m
QA₃ : d₂ = 0.02 m
B_net = B_P + B_Q
Where, B_P is the magnetic field due to wire P and B_Q is the magnetic field due to wire Q.



∴ B_net = 2× 10^-4 T
Hence, magnetic field at A₃ is 2× 10^-4 T
At A₄:
The resultant magnetic field at A₄ would be
From Pythagoras theorem,
(PA₄)² = (PA₃)² + (A₃A₄)²
∴ (PA₄)² = 0.02² + 0.02²
∴ PA₄ = 0.028 m = d₁
Now,

∴B_P = 7.14 × 10^-5 T
Similarly,
QA₄ = 0.028 m = d₂
Thus,


∴ B_Q = 7.14 × 10^-5 T
Substituting to get net magnetic field at A₄,

∴ B_net = 1× 10^-4 T
Hence magnetic field at A₄ is 1× 10^-4 T.

Given:
Current in the equal wires: i= 10 A
Distance between the two wires: a = 2 cm = 0.02 m
Distance between the wire and the point : d = 2 cm = 0.02 m

From the diagram l₁ and l₂ are the wires coming out of the plane.
These two wires and point P form an equilateral triangle of side 0.02 m.
By right hand rule, Magnetic field at P due to l₁ is shown by arrow B₁ tangent to the field’s path and Magnetic field at P due to l₂ is shown by B₂ tangent to the field’s path.
By geometry we can calculate the angle between B₁ and B₂ which turns out to be 60° .
Formula used:
By Ampere’s Law for a current carrying wire is
Where,

B is the magnitude of magnetic field,

μ₀ is the permeability of free space, μ₀= 4π × 10^-7 T mA^-1

d is the distance between the current carrying wire and the required point.
Magnetic field due to wire l₁:

∴ B₁ = 1 × 10^-4 T
Magnetic Field due to l₂ :
∴ B₂ = 1 × 10^-4 T
B₁=B₂ as same magnitude of current is flowing through both the wires and point P is located at same distance from each wires.
Now, angle between and : θ = 60°
Resultant of two vectors is given as
B_R is the resultant magnetic field at P.

∴ B_R = 1.73 × 10^-4 T
Hence, resultant magnetic field at a point 2 cm away from due to two current carrying wires is 1.73 × 10^-4 T.

Given:
Current in both the wires : I = 5 A.

Wires 1 and 2 generate magnetic fields B₁ and B₂ respectively.
By right hand rule, direction of magnetic field in each quadrant due to wires 1 and is shown by X and O.
X : Field is going into the plane
O : Field is coming out of the plane.
Formula used:
By Ampere’s Law for a current carrying wire is
Where,

B is the magnitude of magnetic field,

μ₀ is the permeability of free space and μ₀= 4π × 10^-7 T mA^-1 d is the distance between the current carrying wire and the required point.
(a) At (1m,1m)
As we can see from the diagram at (1,1) the magnetic fields due to wire 1 and 2 are in opposite direction. (X and O)
Hence, net magnetic field would be zero.
(b) At (-1m , 1m)
Here, magnetic field due to wire 1 and 2 would add up as direction of magnetic field is same. ( O and O)
Thus B = B_X + B_Y
B_X is the magnetic field due to wire on x axis
B_Y is the magnetic field due to wire on y axis

∴ B = 2× 10^-6 T
This B would be along z-Axis.
(c) At (-1m,-1m)
As we can see from the diagram at (-1,-1) the magnetic fields due to wire 1 and 2 are in opposite direction. (O and X)
Hence, net magnetic field would be zero.
(d) At (1m,-1m)
Here, magnetic field due to wire 1 and 2 would add up as direction of magnetic field is same. ( X and X)
Thus B = B_X + B_Y
B_X is the magnetic field due to wire on x axis
B_Y is the magnetic field due to wire on y axis