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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 = B _{0}k**

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.

So, verifies answer (d).

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€™.

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).

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).

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.

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.

So answer (b) is verified.

**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 F_{max}= 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°, F_{max} = 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 F_{net} =qE + q(v x B).

(i) The magnetic forces (F_{m} =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 (F_{e} = 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 F_{net} =qE + q(v x B).

Force experience by the charged particle due to electric field F_{e} = qE

Force experience by the charged particle due to magnetic field, F_{m} = 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 F_{net} = 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.

**VERY SHORT ANSWER TYPE QUESTIONS**

**Q.12. Verify that the cyclotron frequency Ï‰ = eB/m has the correct dimensions of [T] ^{â€“1}.**

Here, Î¸ = 90Â°as Î¸ is angle between

so dimensions of below must be equal.

So

So dimensions of Ï‰ is [T

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.

The net acceleration which a rising from this is however, frame independent for inertial frames (non-relativistic physics).

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.

The direction of magnetic field at O

As wire of current I

Hence, force on O

**SHORT ANSWER TYPE QUESTIONS**

**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 +B_{1 }B_{2} andB_{3} 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 B_{1}, B_{2} and B_{3} respectively.

So net magnetic field at origin due to three current-carrying loops B=B_{1}+B_{2}+B_{3}_{}

The resultant of B_{1},B_{2} and B_{3} will be diagonal OR of cube of side B_{1}, B_{2}, B_{3} 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}.**

we know that Centripetal force = .......(2)

By eq. (1) and (2)

Let us consider a magnetic field B = B

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.

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 dl

So, the direction of magnetic field at dl

The direction of magnetic force exerted at dl

F-i(l X B), i.e., F||(i x k) or along â€“ j direction.

Therefore, force due to dl[ on dl

Now, for the direction of magnetic field, at dx, located at (0, 0, 0) due to wire d

So, the magnetic field at dx does not exist.

Force due to dl

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.

maximum current deflection (I

For 2V, R

I

1 x 10

R

For 20V, R

I

10

R

R

For 200V, Rg, R

I

10

20000+R

R

=180000 Î© =180k Î©

Hence, R

length of wire PQ (L

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).

F

also the magnetic force due to a wire

F

The angle between B

ILB=mg (sin 90=0)

B

=

= 51 x 10

= 51 x 10

**LONG ANSWER TYPE QUESTIONS**

**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^{-2}m).

âˆ´ Force acting on arm CD=F_{m} =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 F_{m}=0.98 x 10^{-2} N

Now let the weight mg is added on other side of beam balance to balance the coil

Mg= F_{m}

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 V _{0}. 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.**

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.

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 C

The coordinates of C

The coordinates of C

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 C

As, d has to be greater than

â‡’

Magnetic moment is m = nlA.

Since, the same wire is used in three cases with same potentials, therefore, same current flows in three cases.

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´

A = a

No. of loops n =

Magnetic moment of the coils m = nIA = 3I(a

No. of loops n =

Area,

Magnetic moment of the coils m = nIA

â‡’

â‡’ m is in a geometric series.

Contribution from large distance â†’ 0.

(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.sec

(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)

S(L)

By using arguments as in (b) part, Bz does not depend on length of wire

Magnetic field due to circular or square loops remain same, i.e.

(Bz)

So,(âˆž)

Hence proved.

For 10mA â†’ I

For 100mA â†’ I

For 1 A â†’ I

I

10

= 10 = (10-1) (S

= 10 = 9(S

10

= 10+S

= 10+S

10

= 10+S

= 10+S

10+S1 = 99(S

10 = 9(S

[from (iv)]

[from (v)]

S

So [from (vi)]

Or

S

(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?

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

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

But from (a) part B

Or

R = - B

i.e. direction of resultant is opposite to

R = B

âˆµ

âˆ´ net induced magnetic field at O becomes

- B + B + B + B + B = 3B

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