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# NCERT Exemplars - Moving Charges and Magnetism Notes | EduRev

## Class 12 : NCERT Exemplars - Moving Charges and Magnetism Notes | EduRev

The document NCERT Exemplars - Moving Charges and Magnetism Notes | EduRev is a part of the Class 12 Course Physics Class 12.
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MULTIPLE CHOICE QUESTIONS - I

Q.1. Two charged particles traverse identical helical paths in a completely opposite sense in a uniform magnetic field B = B0k
(a) They have equal z-components of momenta.
(b) They must have equal charges.
(c) They necessarily represent a particle-antiparticle pair.
(d) The charge to mass ratio satisfy:
Ans. (d)
Solution.
For a given pitch,
is angle of velocity of charge particle with X-axis]
If motion is not helical, Î¸ = 0
As path both the particles is identical and helical but of opposite direction in same magnetic field so by law of conservation of momenta.

Q.2. Biot-Savart law indicates that the moving electrons (velocity v) produce a magnetic field B such that
(a) B âŠ¥ v.
(b) B||v.
(c) it obeys inverse cube law.
(d) it is along the line joining the electron and point of observation.
Ans. (a)
Solution.
By Biot-Savart law,
Or
acc to Biot-Savart law, if magnetic field is not perpendicular to the motion of charge then it will not move in helical path, which is not possible for motion of a charge in magnetic field.
So the magnetic field is perpendicular to the direction of flow of charge verifies answer â€˜aâ€™.

Q.3. A current carrying circular loop of radius R is placed in the x-y plane with centre at the origin. Half of the loop with x > 0 is now bent so that it now lies in the y-z plane.
(a) The magnitude of magnetic moment now diminishes.
(b) The magnetic moment does not change.
(c) The magnitude of B at (0.0.z), z >>R increases.
(d) The magnitude of B at (0.0.z), z >>R is unchanged.
Ans. (a)
Solution.
Acc. to Fleming's left hand rule the direction of magnetic field due to current carrying circular loop is perpendicular and it is perpendicular to plane of loop and unidirectional.
In first case, direction of magnetic field is only in positive x-z direction but when it is bent then B due to half loop is along -x axis (due to unfolded loop) and B due to other half loop is along +x direction and equal in magnitude so vector sum of equal and opposite B cancel out each other. Verifies answer (a).

Q.4. An electron is projected with uniform velocity along the axis of a current carrying long solenoid. Which of the following is true?
(a) The electron will be accelerated along the axis.
(b) The electron path will be circular about the axis.
(c) The electron will experience a force at 45Â° to the axis and hence execute a helical path.
(d) The electron will continue to move with uniform velocity along the axis of the solenoid.
Ans. (d)
Solution.
F=q(v x B),
F=qvB x sin Î¸
As the charge is moving in the direction of electric field hence Î¸=0, so force due to E.F. is zero, so will not affect the velocity of moving charge particle. So verifies answer (d).

Q.5. In a cyclotron, a charged particle
(a) undergoes acceleration all the time.
(b) speeds up between the dees because of the magnetic field.
(c) speeds up  in a dee.
(d) slows down within a dee and speeds up between dees.
Ans. (a)
Solution.
Cyclotron is a device used to accelerate positively charged particles (like a-particles, deuterons etc.)
It is based on the fact that the electric field accelerates a charged particle and the perpendicular magnetic field keeps it revolving in circular orbits of constant frequency. Thus a small potential difference would impart enormously large velocities if the particle is made to traverse the potential difference a number of times.

Q.6. A circular current loop of magnetic moment M is in an arbitrary orientation in an external magnetic field B. The work done to rotate the loop by 30Â° about an axis perpendicular to its plane is
(a) MB.
(b)
(c)
(d) zero.
Ans. (b) and (d)
Solution.
when the axis of rotation of loop is along B then angle between  and is 90Â° always. So WD by loop to rotate i.e., WD=MB cos 90Â°. So WD is zero. Verifies option (d).
But when the axis of rotation of loop is not along the direction of B, then direction of vector B and A will change with time.
Work done by loop during orientation in uniform magnetic field.

MULTIPLE CHOICE QUESTIONS - II

Q.7. The gyro-magnetic ratio of an electron in an H-atom, according to Bohr model, is
(a) independent of which orbit it is in.
(b) negative.
(c) positive.
(d) increases with the quantum number n.
Ans. (b)
Solution.
Angular momentum of electron(L)=nh/2Ï€
Magnetic moment of electron M=n[eh/4Ï€m]
gyromagnetic ratio =magnetic moment/angular momentum
2Ï€neh/4Ï€nhm
=e/2m
=constant
It is independent of the orbit in which electron is revolving.
Since e is negative , the geomagnetic ratio is negative. verifies the answer is (b)

Q.8. Consider a wire carrying a steady current, I placed in a uniform magnetic field B perpendicular to its length. Consider the charges inside the wire. It is known that magnetic forces do no work. This implies that,
(a) motion of charges inside the conductor is unaffected by B since they do not absorb energy.
(b) some charges inside the wire move to the surface as a result of B.
(c) if the wire moves under the influence of B , no work is done by the force.
(d) if the wire moves under the influence of B , no work is done by the magnetic force on the ions, assumed fixed  within the wire.
Ans. (b, d)
Solution.
Key concept: If a current carrying straight conductor (length l) is placed in a uniform magnetic field (B) such that it makes an angle Î¸ with the direction of field, then force experienced by it is Fmax= Bil sin Î¸. Direction of this force is obtained by right hand palm rule.
Right-hand palm rule: Stretch the fingers and thumb of right hand at right angles to each other. Then if the fingers point in the direction of field B and thumb in the direction of current z, then normal to the palm will point in the direction of force

If conductor is placed perpendicular to magnetic field, then Î¸ = 90Â°, Fmax = bil

Q.9. Two identical current carrying coaxial loops, carry current I in an opposite sense. A simple amperian loop passes through both of them once. Calling the loop as C,
(a)
(b) the value of is independent of sense of C.
(c) there may be a point on C where B and dl are perpendicular.
(d) B vanishes everywhere on C.
Ans. (b) and (c)
Solution.
Loops are identical placed coaxially and carrying same current in opposite sense. So inside amperian loop of any type direction of current will be opposite by Ampereâ€™s circuital law

As the magnetic field inside (over everywhere) the loop is perpendicular to the direction of plane of loop, so

So, answer (b) and (c) are verified.

Q.10. A cubical region of space is filled with some uniform electric and magnetic fields. An electron enters the cube across one of its faces with velocity v and a positron enters via opposite face with velocity - v. At this instant,
(a) the electric forces on both the particles cause identical accelerations.
(b) the magnetic forces on both the particles cause equal accelerations.
(c) both particles gain or loose energy at the same rate.
(d) the motion of the centre of mass (CM) is determined by B alone.
Ans. (b, c, d)
Solution.
Key concept: This problem is based upon the single moving charge placed with some uniform electric and magnetic fields in space. Then they experiences a force called Lorentz force is given by the relation Fnet  =qE + q(v x B).
(i) The magnetic forces (Fm =q(v x B)), on charge particle is either zero or Fm is perpendicular to v (or component of v) which in turn revolves particles on circular path with uniform speed. In both the cases particles have equal accelerations.
(ii) Due to same electric force (Fe = qE) which is in opposite direction (because of sign of charge) both the particles gain or loss energy at the same rate.
(iii) There is no change of the Centre of Mass (CM) of the particles, therefore the motion of the Centre of Mass (CM) is determined by B alone.

Q.11. A charged particle would continue to move with a constant velocity in a region wherein,
(a) E = 0, B â‰  0.
(b) E â‰  0, B â‰  0.
(c) E â‰  0, B = 0.
(d) E = 0, B = 0.
Ans. (a, b, d)
Solution.
Key concept: This problem is based upon the single moving charge placed with some uniform electric and magnetic fields in space. Then they experiences a force called Lorentz force is given by the relation Fnet =qE + q(v x B).
Force experience by the charged particle due to electric field Fe = qE
Force experience by the charged particle due to magnetic field, Fm = q(v x B)
According to the problem, particle is moving with constant velocity means acceleration of particle is zero and also it is not changing its direction of motion.
This will happen when net force on particle is zero.
(i) if E = 0, and v || B, then Fnet = 0.
(ii) if E â‰  0, B â‰  0 and E, v and B are mutually perpendicular.
And (iii) when both E and B are absent.

Q.12. Verify that the cyclotron frequency Ï‰ = eB/m has the correct dimensions of [T]â€“1.
Ans. In cyclotron, charged particles are placed perpendicular to both E and B. Electric field exert force on particle resulting increase in velocity of particle and magnetic field keeps it into circular path. So it provides the centripetal force for revolution. So,
Here, Î¸ = 90Â°as Î¸ is angle between

so dimensions of below must be equal.
So

So dimensions of Ï‰ is [T-1].

Q.13. Show that a force that does no work must be a velocity dependent force.
Ans. To show that a force that does no work must be a velocity dependent force, then we have to assume that work done by force is zero. As shown by the equation below:

We can write,

So we can say that force F must be velocity dependent, this implies that angle between F and v is 90Â°. If the direction of velocity changes, then direction of force will also change.

Q.14. The magnetic force depends on v which depends on the inertial frame of reference. Does then the magnetic force differ from inertial frame to frame? Is it reasonable that the net acceleration has a different value in different frames of reference?
Ans. As F = q(v x B),velocity depends on frame of reference. Hence The magnetic force is frame dependent. So, yes the magnetic force differ from inertial frame to frame.
The net acceleration which a rising from this is however, frame independent for inertial frames (non-relativistic physics).

Q.15. Describe the motion of a charged particle in a cyclotron if the frequency of the radio frequency (rf) field were doubled.
Ans. The frequency va of the applied voltage (radio frequency) is adjusted so that the polarity of the dees is reversed in the same time that it takes the ions to complete one half of the revolution. The requirement va = vc is called the resonance condition.
When the frequency of the radio frequency (rf) field were doubled, then the resonance condition are violated and the time period of the radio frequency (rf) field were halved. Therefore, the duration in which particle completes half revolution inside the dees, radio frequency completes the cycle.
So, particle will accelerate and decelerate alternatively. So, the radius of path in the dees will remain same.

Q.16. Two long wires carrying current I1 and I2 are arranged as shown in Fig. 4.1. The one carrying current I1 is along is the x-axis. The other carrying current I2 is along a line parallel to the y-axis given by x = 0 and z = d. Find the force exerted at O2 because of the wire along the x-axis.
Ans. We know that force on current (I) carrying conductor placed in magnetic field B is F=I(LxB) =ILB sinÎ¸.
The direction of magnetic field at O2 due to the current I1 is parallel to Y-axis and in -Y direction.
As wire of current I2 is parallel to Y-axis, current in I2 is also along Y-axis. So I2 and B1 (magnetic field due to current I1) are also along Y-axis i.e., angle between I2 and B1 is zero. So magnetic force F2 on wire of current I2 is F2= B1I2 L1 sin 0Â° =0.
Hence, force on O2 due to wire of current I1 is zero.

Q.17. A current carrying loop consists of 3 identical quarter circles of radius R, lying in the positive quadrants of the x-y, y-z and z-x planes with their centres at the origin, joined together. Find the direction and magnitude of B at the origin.
Ans. Consider in figure, 3 quadrants of conductors AB, BC and CD along positive X-Y, Y-Z and Z-X planes respectively. A and D are connected to a battery which is responsible to flow current I through the three quadrants of radius R coordinate of A or D (R,0,0), B(0, R,0) and of C(0,0,R). Now the direction of magnetic field by right-hand thumb rule due to quadrants AB, BC and CD are +BB2 andB3 along +Z, +X and +Y directions respectively. So, at the centre of quadrant

So M.F. due to quadrants AB, BC and CD at their centre O are B1, B2 and B3 respectively.

So net magnetic field at origin due to three current-carrying loops B=B1+B2+B3

The resultant of B1,B2 and B3 will be diagonal OR of cube of side B1, B2, B3 as the

Q.18. A charged particle of charge e and mass m is moving in an electric field E and magnetic field B. Construct dimensionless quantities and quantities of dimension [T ]â€“1.
Ans. When a charged particle is placed in an electric and magnetic field, its starts moving in helical motion where electric field E and B is perpendicular to direction of motion, electric field exert force of charged particle and centripetal force is applied by magnetic force Fm=qvB sin 90Â° =qvB......(1)
we know that Centripetal force = .......(2)
By eq. (1) and (2)

Q.19. An electron enters with a velocity v = v0i into a cubical region (faces parallel to coordinate planes) in which there are uniform electric and magnetic fields. The orbit of the electron is found to spiral down inside the cube in plane parallel to the x-y plane. Suggest a configuration of fields E and B that can lead to it.
Ans. Key concept: Due to magnetic force charge particle revolves in uniform circular motion in x-y plane and due to electric field charge particle increases the speed along x-direction, which in turn increases the radius of circular path and hence, particle traversed on spiral path.
Let us consider a magnetic field B = B0 present in the region and an electron enters with a velocity into cubical region (faces parallel to coordinate planes). The force on electron, using magnetic Lorentz force, is given by

which revolves the electron in x-y plane.
The electric force  accelerates e along z-axis which in turn increases the radius of circular path and hence particle traversed on spiral path.

Q.20. Do magnetic forces obey Newtonâ€™s third law. Verify for two current elements  located at the origin and  located at (0, R, 0). Both carry current I.
Ans. Key concept: In this problem first we have to find the direction of magnetic field due to one wire at the point on other wire, then the magnetic force on that current carrying wire.
According to Biot-Savartâ€™s law, magnetic field B is parallel to idl x r and idl is the current carrying element having its direction along the direction of flow of current.
Here, for the direction of magnetic field, at dl2, located at (0, R, 0) due to wire dlx is given by B || idl x r or i xj (because point (0, R, 0) lies on y-axis), but i x j = k.
So, the direction of magnetic field at dl2 is along z-direction.
The direction of magnetic force exerted at dl2 due to the magnetic field of first wire is along the x-axis.
F-i(l X B), i.e., F||(i x k) or along â€“ j direction.
Therefore, force due to dl[ on dl2 is non-zero.
Now, for the direction of magnetic field, at dx, located at (0, 0, 0) due to wire dis given by B||idl x r or j x â€“ j (because origin lies on y-direction w.r.t. point (0, R, 0), but j x â€“ j = 0.
So, the magnetic field at dx does not exist.
Force due to dl2 on dl1, is zero.
So, magnetic forces do not obey Newtonâ€™s third law. But they obey Newtonâ€™s third law if current carrying element are placed parallel to each other.

Q.21. A multirange voltmeter can be constructed by using a galvanometer circuit as shown in Fig. 4.2. We want to construct a voltmeter that can measure 2V, 20V and 200V using a galvanometer of resistance 10â„¦ and that produces maximum deflection for current of 1 mA. Find R1, R2 and R3 that have to be used.

Ans. Given resistance of Galvanometer(Rg) = 10 Î©,
maximum current deflection (Ig) =1 mA =1 x 10-3A
For 2V, Rg and R1 are connected in series:
Ig(Rg+R1) = 2V
1 x 10-3[10+R1] = 2
For 20V, Rg, R1 and R2 are connected in series:

Ig[Rg+R1+R2]=20V
10-3[10+1990+R2] = 20
R= 20000-2000
For 200V, Rg, R1, R2 and R3 are connected in series:
Ig[Rg+R1+R2+R3] = 200V, as R = (R1+R2)+R3
10-3[10+1990+18000+R3] = 200
20000+R= 200000
R3 = 200000 - 20000

Q.22. A long straight wire carrying current of 25A rests on a table as shown in Fig. 4.3. Another wire PQ of length 1 m, mass 2.5 g carries the same current but in the opposite direction. The wire PQ is free to slide up and down. To what height will PQ rise?
Ans. Given current in wire (I2) = 25 A,
length of wire PQ (L2) =1 m,
mass of wire PQ (m) = 2.5 g
Wire PQ must experience a repulsive force due to magnetic field by wire AB.
Let the wire is balanced at height h thus, magnetic force due to wire AB on PQ should be equal and opposite to weight pf the body (=mg).
Fm=m g.........(1)
also the magnetic force due to a wire
Fm=ILBsinÎ¸.........(2)
The angle between B1 and I in PQ is 90Â°. so by eq. (1) and (2) we get,
ILB=mg    (sin 90=0)
B1I2L2 =mg

=
= 51 x 10-4
= 51 x 10-2cm = 0.51 cm

Q.23. A 100 turn rectangular coil ABCD (in XY plane) is hung from one arm of a balance (Fig. 4.4). A mass 500g is added to the other arm to balance the weight of the coil. A current 4.9 A passes through the coil and a constant magnetic field of 0.2 T acting inward (in xz plane) is switched on such that only arm CD of length 1 cm lies in the field. How much additional mass â€˜mâ€™ must be added to regain the balance?
Ans. The magnetic field is perpendicular to arms BC and AD, so torque will act on CD and AB arms due to it, Coil rotate.
When current of 4.9 A does not pass through the coil the balance measures mass of coil 500g.
On arm AD and BC of rectangular coil magnetic force due to M.F. will be equal and opposite so it will rotate the coil horizontally not vertically up or down. So does not affect the balance.
When current of 4.9 A passes through the coil, downward force acts on arm CD due to magnetic field. Length of arm CD is 1 cm (10-2m).
âˆ´ Force acting on arm CD=Fm =B x I.l = BI1 sin Î¸[Î¸is angle between B and I in CD]
= 0.2 x 4.9 x sin 90Â° x 10-2
Or Fm=0.98 x 10-2 N
Now let the weight mg is added on other side of beam balance to balance the coil
Mg= Fm
m x 9.8=0.98x10-2

Q.24. A rectangular conducting loop consists of two wires on two opposite sides of length l  joined together by rods of length d. The wires are each of the same material but with cross-sections differing by a factor of 2. The thicker wire has a resistance R and the rods are of low resistance, which in turn are connected to a constant voltage source V0. The loop is placed in uniform a magnetic field B at 45Â° to its plane. Find Ï„, the torque exerted by the magnetic field on the loop about an axis through the centres of rods.
Ans. After analyzing the direction of current in both wires, magnetic forces and torques need to be calculated for finding the net torque.

According to the problem, the thicker wire has a resistance R, then the other wire has a resistance 2R as the wires are of the same material so their resistivity remains same.
Now, the force and hence, torque on first wire is given by

Similarly, the force hence torque on other wire is given by

So, net torque,

where A is the area of rectangular coil.

Q.25. An electron and a positron are released from (0, 0, 0) and (0, 0, 1.5R ) respectively, in a uniform magnetic field , each with an equal momentum of magnitude p = e BR. Under what conditions on the direction of momentum will the orbits be non-intersecting circles?
Ans. The magnetic field B is along the x-axis, hence for a circular orbit the momenta of the two particles are in the y-z plane. Let p1 and p2 be the momentum of the electron (eâ€“ ) and positron (e+), respectively. Both traverse a circle of radius R of opposite sense. Let p, make an angle Î¸ with they-axis p2 must make the same angle with y-axis.
The centres of the respective circles must be perpendicular to the momenta and at a distance R. Let the centre of the electron be at Ce and of the positron at Cp.
The coordinates of Ce is Ce = (0, -R sin Î¸, R cos Î¸)
The coordinates of Cp is Cp = [0, -R sin Î¸, (1.5R â€“ R cos Î¸)]

The circular orbits of electron and positron shall not overlap if the distance between the two centers are greater than 2R.
Let d be the distance between Cp and Ce. Then.

As, d has to be greater than
â‡’

Q.26. A uniform conducting wire of length 12a and resistance R is wound up as a current carrying coil in the shape of (i) an equilateral triangle of side a; (ii) a square of sides a and, (iii) a regular hexagon of sides a. The coil is connected to a voltage source V0. Find the magnetic moment of the coils in each case.
Ans. Key concept: In this problem different shapes form figures of different area and the number of loops in each case is different and hence, there magnetic moments varies.
Magnetic moment is m = nlA.
Since, the same wire is used in three cases with same potentials, therefore, same current flows in three cases.
(i) For an equilateral triangle of side a,

As the total wire of length - 12a, so, the no. of loops n =
Magnetic moment of the coils m â€” nIA
As area of triangle is A =

âˆ´
(ii) For a square of sides a,

A = a2
No. of loops n =
Magnetic moment of the coils m = nIA = 3I(a2) = 3Ia2
(iii) For a regular hexagon of sides a,

No. of loops n =
Area,
Magnetic moment of the coils m = nIA
â‡’
â‡’ m is in a geometric series.

Q.27. Consider a circular current-carrying loop of radius R in the x-y plane with centre at origin. Consider the line integral

(a) Show that  (L) monotonically increases with L.
(b) Use an appropriate Amperian loop to show that  ( âˆž )= Âµ 0 I , where I is the current in the wire.
(c) Verify directly the above result.
(d) Suppose we replace the circular coil by a square coil of sides R carrying the same current I. What can you say about  ( L ) and( âˆž ) ?
Ans. (a) B(z) point in the same direction of Z-axis and hence(L) is monotonically function of L as B and dl are along the same direction. So

(b)(L) + Contribution from large distance on contour C=
âˆ´ Asâ†’L âˆž
Contribution from large distance â†’ 0.
âˆ´
(c) The magnetic field due to circular current-carrying loop of radius R in X-Y plane with centre at origin at with centre at origin at any point lying at distance a from origin. Consider a loop of current- carrying conductor placed in X-Y plane. A point P is in +Z direction at distance z, i.e. OP=z.

Again consider an element dz on loop of conductor as shown in figure below.
Let angle between R and QP=Î¸ then magnetic field at P due to loop is
[Integrating both sides w.r.t.z and z can vary from -d to +d]

z=R tan Î¸
Differentiating both sides:
dz=R.sec2 Î¸.dÎ¸

(d) Because the area of square loop is smaller than the area of circular loop, for the same length of conducting wires, hence loop B(z)square loop <loop B(z)circular loop
S(L)sq.= S(L)circular  [âˆµ length of conducting wire are equal]
By using arguments as in (b) part, Bz does not depend on length of wire
âˆ´(âˆž)sq. loop = (âˆž)circular loop
Magnetic field due to circular or square loops remain same, i.e.
(Bz)circular l00p (Bz)Square loop = Î¼0I
So,(âˆž)sq.loop=(âˆž)circular loop = Î¼0I
Hence proved.

Q.28. A multirange current meter can be constructed by using a galvanometer circuit as shown in Fig. 4.5. We want a current meter that can measure 10 mA, 100 mA and 1 A using a galvanometer of resistance 10â„¦ and that produces maximum deflection for current of 1 mA. Find S1, S2 and S3 that have to be used

Ans. We can measure the currents of magnitude 10 mA, 100 mA and 1 A by connecting ammeter A and B, C and D respectively. So,
For 10mA â†’ IgG = (I-Ig) (S1+S2+S3) ...(i)
For 100mA â†’ Ig(G+S1) = (I-Ig)(S2+S3) ...(ii)
For 1 A â†’ Ig(G+S1+S2) =(I-Ig)S3 ...(iii)
I= l mA = 10-3 A and G = 10 Î©
10-3 Î¸10 = (10-2-10-3) (S1+S2+S3) [from (i)]
= 10 = (10-1) (S1+S2+S3)
= 10 = 9(S1+S2+S3) ...(iv)
10-3(10+S1) = (10-1-10-3)(S2+S3) [from (ii)]
= 10+S= (100-1) (S2+S3)
= 10+S= 99(S2+S3) ...(v)
10-3(10+S1+S2) = (1-10-3)(S3) [from (iii)]
= 10+S1+S= (1000-1)S3
= 10+S1+S= 999 S3 ...(vi)
10+S1 = 99(S3+S2) [from (v)]
10 = 9(S3+S2+S1) [from (v)]
[from (iv)]
[from (v)]

So  [from (vi)]
Or

Q.29. Five long wires A, B, C, D and E, each carrying current I are arranged to form edges of a pentagonal prism as shown in Fig. 4.6. Each carries current out of the plane of paper. (a) What will  be magnetic induction at a point on the axis O? Axis is at a distance R from each wire.
(b) What will be the field if current in one of the wires (say A) is switched off?
(c) What if current in one of the wire (say) A is reversed?

Ans. (a) Figure shows that five conductors AAâ€™, BBâ€™, CCâ€™, DDâ€™ and EEâ€™ are along height of regular pentagonal prism ABCDE.

It is given that five identical conducting wires are along the heights of regular pentagon, represented in figure above by AAâ€™, BBâ€™, CCâ€™, DDâ€™, and EEâ€™. Axis of regular pentagon is OOâ€™ will be equidistant (R) from all five conductors, the current is passing through all five conductors are equal let (I).
As the current in all 5 conductors are equal to I and the distance of O from conductors is also equal to R then magnitude of magnetic field due to each conductor will be equal, i.e.

The direction of magnetic field induced can be find out by right hand grip rule, then direction of induced magnetic field at â€˜Oâ€™ due to AAâ€™ will be perpendicular to both AAâ€™ and AO. Angles between any two consecutive magnetic field is
As shown in figures given here

As B = B= B= B= B= B5 and angle between consecutive magnetic fields is 72Â° or symmetric is 360Â° so their resultant at O will be zero, i.e.

Hence, the induced magnetic induction at O due to five conductors as shown in figure is zero.
(b) When current in AAâ€™ is switched off, then B^0 and resultant becomes
R = B+ B+ B+ B5
But from (a) part B+ B+ B+ B+ B5=0
Or
R = - B1

i.e. direction of resultant is opposite to
(c) Here, on reversing the current in AAâ€™, direction of magnetic field due to AAâ€™ becomes
R = B+ B+ B+ B5
âˆµ
âˆ´ net induced magnetic field at O becomes
- B + B + B + B + B = 3B

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