Magnetism and Matter Test 6

(a) The point charge moves in circle as shown in Fig.
The magnetic field vectors at a point P on the axis of circle
are B→AandB→CB→AandB→C at the instant the point charge is at A and C
respectively, as shown in Fig.

 We know that, 

B = μoI/4πa(sinθ2+sinθ1)

The net magnetic field due to wires from left to night and from right to left is zero as they cancel out. By applying the right-hand thumb rule. As the left sides, the magnetic field adds up & direction is upward whereas the right-side magnetic field adds up to but direction is downward. So, cancels up.

And for the wires from up to down or down to up, it is the wires cancel its conjugate according to right-hand thumb-rule.

Thus, net magnetic field at a point is 0. 

Force due to both wires are inward.
=2 x (μ_o​I/4π)[(1/(r-x))+(1/x)]
=(μ_o​I/2π)[(1/(r-x))+(1/x)]

Magnetic field at some x is given by,
[μ₀(4I)/2π(d-x)]k+ [μ₀I/2π(d+x)]k
= μ₀/2π[(4/d-x)+(1/d+x)]
= μ₀/2π[5d+3x/d²-x²]
The correct answer is option C.
 

B₁= μ₀ni/2r₁
B₂= μ₀ni/2r₂
B₁/B₂=r₂/r₁=3/4
r₁:r₂=4/3

Magnetic field at the axial of bar magnet is given by the relation as :
 B=(μ_O/2π)​​(M​/d³)
where d is the distance
 So, 
B∝1/d^3​
​B_A/B_B ​​= d_B​³​/ d_A​³
​BA/ BB​​=27/1​
Hence, the ratio of magnetic field of A and B is 27:1.
 

Magnetic lines of force are continuous.
The tangent at each point of the magnetic field lines the direction of the magnetic field. And we know that magnetic fields exist in each and every point  in the vicinity of the magnet.

Magnetic field at the centre due to current ABC arc is 
B₁​= (μ₀​​/4π)(2πI₁/r)​​(q/2π)    (Directed upwards)
Magnetic field at the centre due to current in ADC arc is
B_2​=μ0​_​2πI₂​​(2π−q)/4π r 2π     (Directed downwards)
Therefore, net magnetic field at the centre
B=[μ0​​₂πI1​​q/ 4πr₂π] −[2πI₂​​(2π−q) /r​ 2π]
Also, I_1​= V/R1​ ​=[V/(ρl₁​/A)]​= VA/ρrq​ and
I₂​= V/R₂​ ​= E/(ρl₂​/A) ​= VA​/ρr(2π−q)
∴B= (μ₀/4π​​)[(VA/ρrq)​×(q/r)​− (VA /ρr(2r−q)) ​×((2π−q)/r)​]=0

The magnetic field due to a current carrying coil at the axis of the coil at a distance x from the centre of coil having N turns is given by,
B= μ₀NIr²/2(r²+x²)^3/2
So, here I=1A, N=10, r=4x10^-2m and x=6x10^-2m
B₁=2π x 10 x 1 x (16x10^-4/16+36)^3/2 x-10^-6)
B₁=2.67x10^-5T=0.264x10-4T
B=μ₀NI/2r so,
B₂=[(2πx10^-7x10x1)/(2x4x10^-2)]T
B₂=1.57x10^-4T
And since the current in both coil are same but in the in the opposite direction so magnetic fields at centre of either coils would be opposite so net magnetic field at the centre of either coil is,
B=B₂-B₁
B=(1.57-0.267)x10^-4T
B=1.303x10^-4T
The magnetic field at the midway of the two centres would be zero because the magnetic field at this point due to both coil is equal but opposite due to opposite current flowing in the coils. thus, B=0.
Hence  option B is correct.
 

Magnetic field at the centre of a coil  Bc​=μo​IN​/2r
where N is the number of turns.
For semi-circular coil N=0.5
So, magnetic field at C due to semi-circular coil of radius R1​,
B1​=​ μo​I(0.5)/2R1​=​μo​I​/4R1  out of plane of paper
So, magnetic field at C due to semi-circular coil of radius R2​,
B2​=​μo​I(0.5)/2R2​=​μo​I​/4R2  into the plane of paper
Magnetic field at C due to section SR and PQ is zero.
So net magnetic field at C   B_net​=B₁​−B₂​=(μo​I/4)​[(1/R₁)​​−(1/R_2​)​]

B=Bouter+Binner
Since the current carried by the outer and inner are equal and in opposite direction,
magnetic field produced by the cylinders at a point P (distance r from central axis) is
Bouter= (μ0/4π)i(dl×r/r3)
Binner=−(μ0/4π)i(dl×r/r3)
∴B=Bouter+Binner=0

Force on positive charges in the rod will be in direction given by qv×B, i.e., towards A.
Hence force on the electron will be in the opposite direction i.e., towards B. So the negative charge will move towards B.
Therefore, B will be negatively charged and A will be positively charged.

F(m) = qvB sin(90°)
Applied potential difference "V" = 12 kV = 12 × 10³ V
Speed of a charged particle, v =1.0 × 10⁶ m s⁻¹
Magnetic field strength "B" = 0.2 T
As per the question, a charged particle is injected perpendicularly into the magnetic field.
We know:
qV = (½)mv²
=>m / q = (½)mv²
= (2 × 12×10³)/ (1×10⁶)²
= 24×10⁻⁹
r = mv / qB
r=(24×10⁻⁹ × 10⁶) / 0.2
r = 12 × 10⁻² m
= 12 cm
Thus, the radius of the circle is 12 cm.
 

Path C is undeviated. Therefore, it is of neutron's path. From Fleming's left-hand rule magnitude force on positive charge will be leftwards and on negative charge is rightwards. Therefore, tack D is of electron. Among A and B one of proton and other of α−particle ≤.
Further, r=mv/Bq or r∝m/q
Since, (m/q)>(m/q)_P
∴r_α>r_P
or track B is of α particle.
 

v_x​=vcosθ
v_y​=vsinθ
Since v_y​≠0, this component will be responsible for motion along the direction B( i.e y−axis).
The path will be helical and pitch will be constant.
Radius r of the circle perpendicular to the magnetic field B
r=mvcosθ/qB​=1.6×10⁻²⁷×(2×10⁶×cos60⁰)/ 1.6×10⁻¹⁹×0.1​=0.1m
T=2πr/v​=2×π×0.1/2×10⁶×cos600 ​=2π×10⁻⁷secs

Radius    r= √2mK​​/qB
⟹ r∝ √m/q​​
Because the deflection is inversely proportional radius so,
  q/√m​=1/1​ : 1/√4 ​​: 2/√16​=1:1/2​:1/2​
Therefore, H+ deflects most and He+ and O+2 will be equally deflected.

A permeable substance is one through which the magnetic lines of force can pass through easily.

mv = BQr
or r = (m/BQ) v
As m, Q and B are the same for all the electrons, r ∝ v. 

Pitch =Vcosθ×T
=Vcosθ×(2πm/qB)
Where m=1kg, q=1C, B=1T, V=1 m/s
=√3π .

Magnetic field is in -Kdirection so direction of force.

So the answer is y axis.
 

A charged particle moves in a circle when its velocity is perpendicular to the magnetic field. When it forms an acute angle with the magnetic field, it can be resolved in two components, parallel and perpendicular. The perpendicular components tends to move it in circle, the parallel components tends it to move along the magnetic field to form a helical motion of uniform radius and pitch.

Let the velocity of the electron is v

Then

If the electric field is E then

Lorentz Force

or,

For force to be zero

or,

To get the RHS in the direction of , B should be acting in the direction of  . As we know that

Therefore the direction of the magnetic field must be in the negative z-axis.

When the particle is released from rest, it starts to move along the fields due to application of electric field. 
The magnetic force acting on a moving charged particle is given by 

Since velocity and magnetic fields are parallel, no magnetic force acts on the particle, hence it continues to move in a straight line along with constant acceleration.