Test: Magnetism & Matter - 3 - NEET MCQ

As point O is along the length of segments L and M, so the field at O due to these segments will be zero. 
Also, point O is near one end of a long wire.
The resultant field at O, B_R​=B_P​+B_Q​, this field will be into the plane paper.
⇒B_R​= μ₀I/4π​​RO​+​​ μ₀I/4πSO​
But, RO=SO=0.02m
Hence, BR​=2× (μ₀/4π) ​​×(10/0.02)=2×10⁻⁷ (10/0.02) ​=10⁻⁴Wb m⁻²

We know that,
B₁=B₂μ₀/2πr
Angle between B₁ and B₂
BR= √[B₁²+B₂²+2B₁B₂Cos(180-2θ)]
= √2B²+2B²(-Cos2θ) 
=B√[2(1-Cos²θ)]
=B√[2(sin²θ)]
=2Bsina
B=(2μ₀i/2πr) x (a/r)
B=(2μ₀ia/2πr²)

Currents are in the same direction, so at the middle B=D
And the B varies exponentially not linearly,
So,

Magnetic field at P is equal to the magnetic field due to AB,CD,EF and so on.
 Magnetic field due to AB is +Z direction.
and CD is negative −Z direction and EP is positive +Zdrection and so on.
B_P​= (​μ₀​I/4πd₁)​ [sin30^°−sin(−30^°)k]+ ​(μ₀​I/4πd₂)​[sin30°−sin(−30°)l−k]+ μ0​I /4πd₃​ ​[sin30°−sin(−30°)k]+...∞
B_P​=2sin30°× (μ₀​I​/4π) [(1/d₁)​​−(1/d₂​)​+(1/d_3​)​−(1/d₄​)​+(1/d₅​)​....∞]k
= μ0​I/4π ​[(1/acos30°)​−(1/2acos30°)​+(1/3acos30°)−....∞]k
= μ_0​I/4π(acos30°) ​[1−(1/2)​+(1/3)​−(1/4)​+(1/5)....∞]k
=(2μ_0​I/4π√​3​a) ​ln(1+1)k= (μ₀​I/4π√3​a) ​ln2²k=(μ₀​I/4π√3​a) ​ln4k

Magnetic field in X-Y plane at position (x,y) due to both wires must be zero
current carrying wire with current i1 is taken along X axis
now the magnetic field at the given point is
B₁= μ₀i₁/2πy
now at the same point the magnetic field due to other wire placed along Y axis is given by,
B₂= μ₀i₂/2πx
now we have to make the magnitude of magnetic field due to both wires same
so, B₁=B₂
μ₀i₁/2πy= μ₀i₂/2πx
now by solving above we have,
y=(i₁/i₂)x
so it’s a straight line with slope i₁/i₂ and passing from origin

If we take plane parallel to the length of wire function the all prints one plane will be current Field will be inversely proportional to the perpendicular distance from the wire

As E=Bv, thus X has a dimension of velocity.
Also, c=1/μo​ϵo​​, where c is the speed of light.
So Y must have a dimension of velocity.
Now Z=l/CR​=l×(V/Q)​×(I/V)​=l×(1/Q)​×Q/t​=l/t​
Thus, Z also has a dimension of velocity.

On x-axis
A.  B=(μ0I/2πa)- (μ0I/2πa)=0
B.  On y axis say at (y, 0, 0)
B=[-μ0I/2π(a+y)]k + [μ0I/2π(a-y)]k
So except at origin, B has only z-components.
B cannot have x-component as B is perpendicular to direction of I. 
 

According to all figure given,
The magnetic field due to square shape in all options will be zero, Because distribution of current will be less in the ϕ longer path. But its cumulative effect of the magnetic field at the centre will be the same.
 
In option (A) and (B) current line is passing through the centre, So its perpendicular distance will be zero, so magnetic field in Both (A) and (B) will be zero due to incoming and outgoing current.
Refer image .1
 
Whereas in case of (D) Magnetic field due to incoming current and outgoing current will be equal and opposite so it will cancel out these effects.
Refer image .2
be in case of (C) Magnetic field will be in same direction with equal magnitude by the incoming and outgoing current, So its effect will not cancel 
Option C.

Current around a long straight wire from Ampere's Law is given by-
B(2πr)=μ₀​i
⟹B=μ₀​i​/3πr
r is the distance of the point from the axis of the wire=√ (1²+1²​)=√2​
⟹B=μ_0​i/2√2​π ​
The magnetic field is perpendicular to the vector pointing from the axis to the point at which the magnetic field is found.
The vector points from (1,0,0) to (1,1,1). Thus it makes angle 45^oC with the xy plane.

So, three rings each having equal radius R, are placed mutually perpendicular to each other and each having its centre at the origin of coordinate system If current I is flowing through each ring them the magnitude of the magnetic field at the common centre is:
Now,
we know,
B for circular loop = ui/2r. This formula is for the magnetic field at the centre of a circular loop of current.
So,

We know that,
I=V₀nAe
V₀=i/2ae

F=q(VxB)
F=qVBsin θ                 (Here, θ=90^o ,sinθ=1)
F=qVB                        (Here, V=I/nAe- drift velocity.)
F=ex (I/nAe) x B
F=IB/nA
 

∫B.dl=µ₀I(total current passing through the closed loop)
Currents outside the loop don’t contribute anything to the loop so I5 can be neglected, Also I₄ and I₆ are going into the loop.
=>∫B.dl= µ0(I₁+I₂+I₃-I₄-I₆)
  = µ₀i wb/m

I=1A, x=6x10^-2  r=4x10-2m, N=10
B₁= μ₀​NIr2​/2(r²+x²)^3/2 
B₁=2πx10^-7x10x16x10^-4/(16x10^-4+36x10^-4)^3/2
B₁=32x3.14x10^-10/(52)^3/2x10^-6
B₁=2.67x10-5T
Magnetic field due to another coil at centre is,
B₂= μ₀​NI/2r=2πx10^-7x10/4x10^-2
Since both are in opposite direction,
B=B₂-B₁
  =(1.57-0.267)x10^-4
B=1.3x10^-4T

We know Magnetic field at centre of a coil having Current I, radius r and number of turns N is given as B=μ₀​NI​ /2r
and at a distance x from the centre and on the axis is B= [μ_0​Nix] /[2(x2+r2)] ​ and its direction is given by right-hand screw rule i.e.
 if the observer sees that the current is anticlockwise then the field will be towards the observer and vice-versa, as shown in the figure−1 
 
Part a 
Field at center O1​  will be difference of field due to loop-1 and loop-2 
At O₁​  B_net​ =  (μ_0​NI​/2r)  - [μ0​Nix]/[2(x²+r²)] ​ towards observer 
putting μ₀​=4π×10⁻⁷ ,N=6, r=.04 meter ,x= .06 meter and I = 1 all units in S.I. system
we get B_net​=5.07×10⁻⁵ Tesla towards the observer
Similarly, At O₁​ the filed due to circle 2 will dominate so net filed will be 
B_net​=5.07×10⁻⁵ Tesla in a direction away from the observer.
 
Part b 
According to the above theory the field at the midpoint on the axis of joining the centres will be equal and opposite as current in loops is opposite so net field is zero there at O 

 

From the work- energy theorem
W=△K.E.
⇒qE.2a=(1/2)​m[4υ²−υ²]
⇒E=(3/4)​( mυ^2​/qa)
At P rate of work done by E field will be 
=F.υ=(qE)(υ)cos0°
=(3/4)​( mυ³/a​)

From the work- energy theorem
W=△K.E.
⇒qE.2a=(1/2)​m[4υ²−υ²]
⇒E=(3/4)​( mυ²​/qa)
At P rate of work done by E field will be 
=F.υ=(qE)(υ)cos0°
=(3/4)​( mυ³/a​)
At Q rate of work done
=F.υ=(qE)(2υ)cos90°=0
And work done by the magnetic field is always zero.