Q: Two identical magnets, with length l and mass M each are arranged vertically inside a glass tube as shown in the figure. The upper magnet remains suspended in air above the lower one so that the distance between the nearest poles of the magnets is x(<<l). Find the pole strength (m) of the magnets. Assume that the magnetic length, for each magnet, is same as the geometric length.

First, let's write the force due to lower magnet on the north pole of the upper magnet. Force due to south pole of the lower magnet can be neglected compared to the force applied by the north pole (∵l>>x) ∴FN=4πμ0x2m⋅m(↑) Force on the south pole of the upper magnet due to the lower magnet can be written as FS=4πμ0l2m⋅m−4πμ0(2l)2m⋅m=434πμ0l2m⋅m(↓) For equilibrium FN=FS+Mg or, 4πμ0x2m2=434πμ0l2m2+Mg or, 4πμ0m2[x21−43l21]=Mg Again, x<<, permits us to neglect the second term inside the bracket. In effect, we are saying that FS<<FN. ⇒4πμ0x2m2=Mg⇒m=x⋅μ04πMg

Q: A particle having mass m and charge q is projected from the origin with a velocity v=v0(i^+j^). The region has a uniform magnetic field B=B0k^. Find the x and y component of the particle's velocity as a function of time assuming that it was projected at time t=0.

The circular path of the particle is as shown in figure. The speed of the particle is 2v0 and its initial velocity is inclined at 45∘ to the x axis. In time t the particle rotates by an angle θ to reach point A and its velocity now makes an angle 45∘−θ with the x direction. Recall that the speed does not change. ∴vx=2v0cos(4π−ωt)=2v0cos(4π−mqBt) vy=2v0sin(4π−ωt)=2v0sin(4π−mqBt)

Q: A particle having mass m and charge q is projected from the origin with a velocity v=v0i^ in a non uniform magnetic field B=−(B0x)k^, Here, v0 and B0 are positive constants of proper dimensions. Find the maximum positive x-coordinate of the particle during the motion.

The particle starts at O, moves along a curved path and its velocity at Q becomes parallel to y direction. The x-coordinate (x0) for Q is our required answer. P is an intermediate point having coordinates (x,y). The magnetic force on the particle at P is F=q(vxi^+vyj^)×(−B0xk^)=qB0vxxj^−qB0vyxi^∴mdtdvx=−qB0vyx and mdtdvy=qB0vx⋅x or, dxdvy⋅dtdx=mqB0vx⋅x or, dxdvy⋅vx=mqB0vx⋅x or, dxdvy=mqB0⋅x⇒dvy=mqB0xdx or, ∫0v0dvy=mqB0∫0x0xdx or, v0=mqB02x02 ⇒x0=qB02mv0

Q: A vertically downward magnetic field B=Bj^ exists in a region. A particle having charge q and mass m is projected with velocity v=v0i^ from a point that is at a height h above the ground. Acceleration due to gravity is g. (a) Write the x,y and z components of velocity of the particle after time t. (b) Take the point of projection as the origin and write the coordinates of the particle at time t. (c) In how much time will the particle hit the gorund?

First consider the motion in horizontal xz plane. Angular speed, ω=mqB Angular displacement in time t θ=ωt=mqBt (a) vx=v0cosθ=v0cos(mqBt)vz=v0sinθ=v0sin(mqBt) In y direction (the vertically downward direction), the particle moves with a constant acceleration g. ∴vy=gt (b) x=Rsinθ=qBmv0sin(mqBt) z=R−Rcosθ=R[1−cos(mqBt)]y=21gt2 (c) h=21gt2⇒t=g2h

Q: A ring of radius r carries a current I. It is placed horizontally above the north pole of a magnet. The magnetic field makes an angle θ with the vertical at all points on the circumference of the ring. Strength of the field at circumference of the ring is B. (a) Find the magnetic force on the ring. (b) Find the tension in the ring.

The current loop is in a non uniform field. Net force on it may not be zero. The vertical component of the magnetic field is BV=Bcosθ and its horizontal component is BH=Bsinθ. First we will consider force due to the horizontal field. Force on a current element of length dl due to this component of the field is dF=BHIdlsin90∘ Direction of force on every small element is vertically downward. Resultant downward force is F=BHI∫dl=BHI⋅2πr=2πBIrsinθ Now, consider the effect of vertical component of the field (Bv). Every small current element experiences a radially outward force. The vector sum of all these forces is zero. But a tension is developed in the ring due to this radial pull. From the last example, tension developed in the ring is T=IrBv=IrBcosθ Obviously, the resultant force on the ring is F=2πBIrsinθ

Q: A wire is bent into U shape (ADCB) and is pivoted so that it can swing about the horizontal axis AB. Length of each arm of the wire frame is l. Mass per unit length of the wire is λ. There exists a uniform vertical magnetic field. When a current I is passed through the frame it deflects by an angle θ in its equilibrium position. (a) Find the strength of magnetic field B. (b) Find the force applied by the pivot on the frame in equilibrium position.

(a) To keep the diagram two dimensional, let's draw one arm AD of the frame. The axis AB is perpendicular to the plane of the figure. Assume that you are looking from side BC. AD1 is the equilibrium position. M is mid point of the arm AD. The magnetic force acting on the arm is directed into the plane of this figure. This force is parallel to the axis AB and has no torque about it. Torque due to gravity is τgAD=(λlg)(AMsinθ)=(λlg)(2lsinθ)=21λl2 gsinθ Similarly, torque (about AB ) due to magnetic force on BC1 is zero. Gravitational torque is same as the above. Observer is standing on BC side. He is looking towards BA. Now, let's consider the horizontal arm DC. The diagram shows the cross section of the wire at mid point N. The current is towards the observer when he is looking from B to A. Torque due to gravity TgDC=(λlg)(lsinθ)=λl2 gsinθ Torque due to the magnetic force is (about AB ) τB=FBlcosθ=(IlB)(lcosθ)=Il2Bcosθ. In equilibrium, the magnetic torque balances the gravitational torque Il2Bcosθ=21λl2gsinθ+21λl2gsinθ+λl2gsinθ or, IBcosθ=2λgsinθ or, B=B2λgtanθ (b) In equilibrium, the magnetic force on AD and BC are horizontal and oppositely directed. They cancel out. Magnetic force on DC is horizontal equal to FB=IlB The pivot must apply equal and opposite horizontal force to balance this. Apart from this, the pivot must apply a vertical force to balance the weight of the frame. This force is Fg=3λlg ∴ Resultant force by the pivot is F=FB2+Fg2=I2l2B+9λ2l2g

Q: A conducting wire is bent in U shape. Mass of each of the segments PQ and TR is m and their length is L. Mass of segment QR is 2m and its length is 2L. The conducting frame can rotate freely about the horizontal axis AB. Between P and T there is a capacitor of capacitance C, a cell of emf ε and a switch S. Resistance of the loop PQRTP is nearly zero. A uniform vertical magnetic field B exists in the entire space. The switch S is closed. Find the angular speed acquired by the wire frame a short time after the switch is closed.

Charge flow in the loop is Q=Cε When current in the loop is i, its magnetic dipole moment is Direction of M is out of the plane of the figure. Direction of magnetic torque vector is along BA. τ=MBsin90∘=2L2Bi The current lasts for a very small interval Δt. Impulse of the torque is ∫0Δtτdt=2L2B∫0Δtτdt=2L2BQ=2L2BCε Now, Let's write the moment of inertia of the loop about axis AB. I=3mL2+3mL2+2m(L)2=38mL2 Angular impulse = Change in angular momentum of the loop. ⇒2 L2BCε=38mL2.ω or, ω=43mBCε

Q: A non-conducting rigid rod of length L has particles attached to its ends. The particles have charge +q and −q on them and each has mass m. The rod is moving in mass m. The rod is moving in xy plane in a space which has a uniform magnetic field Bk^, At a particular instant the rod makes an angle θ with the positive x axis, its centre is moving with velocity ui^ and it is rotating with an angular velocity ωk^. Find the net force and the net torque acting on the rod at this instant.

Consider, only rotation. Velocities of the particles are as shown in figure. Force on +q charge is F=q2ωLB directed along the rod as shown in figure. Same force acts on the other charge. These forces do not produce a torque but have a resultant along the rod. The resultant force is F0=qωLB Now, consider translational motion only. Forces on the two particles are F=quB in the directions shown in figure. Resultant of these two forces is zero but they produce a torque. Force due to translation τ=2F2Lcosθ=(quB)Lcosθ=quBLcosθ. When we superimpose the effects of rotation and translation, the net force and the net torque can be written as Fnet =qωLBτnet =quBLcosθ

Q: A ring has mass m and radius R. A wire around the ring carries a current i. The ring is mounted on a massless horizontal axle. The ring-axle arrangement is suspended using two light threads tied to the ends (see figure). A uniform vertical magnetic field exists in the entire space. Find the tension in the two threads.

Net force on the ring axle arrangement is zero. ∴T1+T2=mg Magnetic torque is τB=i⋅πR2⋅Bsin90∘=iπR2B Direction of the torque is as shown in figure. Torque of T2 (about O ) is also in same direction. Torque due to T1 opposes the resultant of torques T2l and τB. ∴ or, T1l=T2l+iπR2BT1−T2=li⋅πR2B Solving (i) and (ii) gives T1=2mg+2li⋅πR2B and T2=2mg−2li⋅πR2B

Q: A current carrying wire is bent in L shape as shown. Find the direction of magnetic field at point P due to the current in the wire.

Field at P due to straight wire AB as well as due to BC is directed into the plane of the figure.

Question 1

Two identical magnets, with length l and mass M each are arranged vertically inside a glass tube as shown in the figure. The upper magnet remains suspended in air above the lower one so that the distance between the nearest poles of the magnets is x(<<l). Find the pole strength (m) of the magnets. Assume that the magnetic length, for each magnet, is same as the geometric length.

Accepted Answer

First, let's write the force due to lower magnet on the north pole of the upper magnet. Force due to south pole of the lower magnet can be neglected compared to the force applied by the north pole (∵l>>x) ∴FN=4πμ0x2m⋅m(↑) Force on the south pole of the upper magnet due to the lower magnet can be written as FS=4πμ0l2m⋅m−4πμ0(2l)2m⋅m=434πμ0l2m⋅m(↓) For equilibrium FN=FS+Mg or, 4πμ0x2m2=434πμ0l2m2+Mg or, 4πμ0m2[x21−43l21]=Mg Again, x<<, permits us to neglect the second term inside the bracket. In effect, we are saying that FS<

Question 2

A particle having mass m and charge q is projected from the origin with a velocity v=v0​(i^+j^​). The region has a uniform magnetic field B=B0​k^. Find the x and y component of the particle's velocity as a function of time assuming that it was projected at time t=0.

Accepted Answer

The circular path of the particle is as shown in figure. The speed of the particle is 2​v0​ and its initial velocity is inclined at 45∘ to the x axis.
In time t the particle rotates by an angle θ to reach point A and its velocity now makes an angle 45∘−θ with the x direction. Recall that the speed does not change.

∴vx​=2​v0​cos(4π​−ωt)=2​v0​cos(4π​−mqBt​)
vy​=2​v0​sin(4π​−ωt)=2​v0​sin(4π​−mqBt​)

Question 3

A particle having mass m and charge q is projected from the origin with a velocity v=v0​i^ in a non uniform magnetic field B=−(B0​x)k^, Here, v0​ and B0​ are positive constants of proper dimensions. Find the maximum positive x-coordinate of the particle during the motion.

Accepted Answer

The particle starts at O, moves along a curved path and its velocity at Q becomes parallel to y direction. The x-coordinate (x0​) for Q is our required answer.

P is an intermediate point having coordinates (x,y). The magnetic force on the particle at P is
F=q(vx​i^+vy​j^​)×(−B0​xk^)=qB0​vx​xj^​−qB0​vy​xi^∴mdtdvx​​=−qB0​vy​x and mdtdvy​​=qB0​vx​⋅x or, dxdvy​​⋅dtdx​=mqB0​vx​​⋅x​
or, dxdvy​​⋅vx​=mqB0​vx​​⋅x
or, dxdvy​​=mqB0​​⋅x⇒dvy​=mqB0​​xdx
or, ∫0v0​​dvy​=mqB0​​∫0x0​​xdx
or, v0​=mqB0​​2x02​​
⇒x0​=qB0​2mv0​​​

Question 4

A vertically downward magnetic field B=Bj^​ exists in a region. A particle having charge q and mass m is projected with velocity v=v0​i^ from a point that is at a height h above the ground. Acceleration due to gravity is g.
(a) Write the x,y and z components of velocity of the particle after time t.
(b) Take the point of projection as the origin and write the coordinates of the particle at time t.
(c) In how much time will the particle hit the gorund?

Accepted Answer

First consider the motion in horizontal xz plane.
Angular speed, ω=mqB​ Angular displacement in time t
θ=ωt=mqBt​

(a)
vx​=v0​cosθ=v0​cos(mqBt​)vz​=v0​sinθ=v0​sin(mqBt​)​
In y direction (the vertically downward direction), the particle moves with a constant acceleration g.
∴vy​=gt
(b) x=Rsinθ=qBmv0​​sin(mqBt​)
z=R−Rcosθ=R[1−cos(mqBt​)]y=21​gt2​
(c) h=21​gt2⇒t=g2h​​

Question 5

A ring of radius r carries a current I. It is placed horizontally above the north pole of a magnet. The magnetic field makes an angle θ with the vertical at all points on the circumference of the ring. Strength of the field at circumference of the ring is B.

(a) Find the magnetic force on the ring.
(b) Find the tension in the ring.

Accepted Answer

The current loop is in a non uniform field. Net force on it may not be zero.
The vertical component of the magnetic field is BV​=Bcosθ and its horizontal component is BH​=Bsinθ.

First we will consider force due to the horizontal field. Force on a current element of length dl due to this component of the field is
dF=BH​Idlsin90∘
Direction of force on every
small element is vertically
downward. Resultant downward
force is
F​=BH​I∫dl=BH​I⋅2πr=2πBIrsinθ​
Now, consider the effect of vertical component of the field (Bv​). Every small current element experiences a radially outward force. The vector sum of all these forces is zero. But a tension is developed in the ring due to this radial pull. From the last example, tension developed in the ring is
T=IrBv​=IrBcosθ
Obviously, the resultant force on the ring is
F=2πBIrsinθ

Question 6

A wire is bent into U shape (ADCB) and is pivoted so that it can swing about the horizontal axis AB. Length of each arm of the wire frame is l. Mass per unit length of the wire is λ. There exists a uniform vertical magnetic field. When a current I is passed through the frame it deflects by an angle θ in its equilibrium position.

(a) Find the strength of magnetic field B.
(b) Find the force applied by the pivot on the frame in equilibrium position.

Accepted Answer

(a) To keep the diagram two dimensional, let's draw one arm AD of the frame. The axis AB is perpendicular to the plane of the figure. Assume that you are looking from side BC.
AD1​ is the equilibrium position. M is mid point of the arm AD. The magnetic force acting on the arm is directed into the plane of this figure. This force is parallel to the axis AB and has no torque about it.

Torque due to gravity is
τgAD​​=(λlg)(AMsinθ)=(λlg)(2l​sinθ)=21​λl2 gsinθ​
Similarly, torque (about AB ) due to magnetic force on BC1​ is zero. Gravitational torque is same as the above.

Observer is standing on BC side. He is looking towards BA.
Now, let's consider the horizontal arm DC. The diagram shows the cross section of the wire at mid point N. The current is towards the observer when he is looking from B to A.
Torque due to gravity
TgDC​=(λlg)(lsinθ)=λl2 gsinθ
Torque due to the magnetic force is (about AB )
τB​​=FB​lcosθ=(IlB)(lcosθ)=Il2Bcosθ.​
In equilibrium, the magnetic torque balances the gravitational torque
Il2Bcosθ=21​λl2gsinθ+21​λl2gsinθ+λl2gsinθ or, IBcosθ=2λgsinθ
or, B=B2λg​tanθ
(b) In equilibrium, the magnetic force on AD and BC are horizontal and oppositely directed. They cancel out.
Magnetic force on DC is horizontal equal to
FB​=IlB
The pivot must apply equal and opposite horizontal force to balance this. Apart from this, the pivot must apply a vertical force to balance the weight of the frame. This force is
Fg​=3λlg
∴ Resultant force by the pivot is
F=FB2​+Fg2​​=I2l2B+9λ2l2g​

Question 7

A conducting wire is bent in U shape. Mass of each of the segments PQ and TR is m and their length is L. Mass of segment QR is 2m and its length is 2L.

The conducting frame can rotate freely about the horizontal axis AB. Between P and T there is a capacitor of capacitance C, a cell of emf ε and a switch S. Resistance of the loop PQRTP is nearly zero. A uniform vertical magnetic field B exists in the entire space. The switch S is closed. Find the angular speed acquired by the wire frame a short time after the switch is closed.

Accepted Answer

Charge flow in the loop is Q=Cε
When current in the loop is i, its magnetic dipole moment is

Direction of M is out of the plane of the figure. Direction of magnetic torque vector is along BA.
τ=MBsin90∘=2L2Bi
The current lasts for a very small interval Δt. Impulse of the torque is
∫0Δt​τdt​=2L2B∫0Δt​τdt=2L2BQ=2L2BCε​
Now, Let's write the moment of inertia of the loop about axis AB.
I=3mL2​+3mL2​+2m(L)2=38​mL2
Angular impulse = Change in angular momentum of the loop.
⇒2 L2BCε=38​mL2.ω or, ω=43​mBCε​​

Question 8

A non-conducting rigid rod of length L has particles attached to its ends. The particles have charge +q and −q on them and each has mass m. The rod is moving in mass m. The rod is moving in xy plane in a space which has a uniform magnetic field Bk^, At a particular instant the rod makes an angle θ with the positive x axis, its centre is moving with velocity ui^ and it is rotating with an angular velocity ωk^. Find the net force
and the net torque acting on the rod at this instant.

Accepted Answer

Consider, only rotation. Velocities of the particles are as shown in figure. Force on +q charge is F=q2ωL​B directed along the rod as shown in figure. Same force acts on the other charge.
These forces do not produce a torque but have a resultant along the rod. The resultant force is F0​=qωLB

Now, consider translational motion only. Forces on the two particles are F=quB in the directions shown in figure. Resultant of these two forces is zero but they produce a torque.

Force due to translation
τ=2F2L​cosθ=(quB)Lcosθ=quBLcosθ.
When we superimpose the effects of rotation and translation, the net force and the net torque can be written as
Fnet ​=qωLBτnet ​=quBLcosθ​

Question 9

A ring has mass m and radius R. A wire around the ring carries a current i. The ring is mounted on a massless horizontal axle. The ring-axle arrangement is suspended using two light threads tied to the ends (see figure). A uniform vertical magnetic field exists in the entire space. Find the tension in the two threads.

Accepted Answer

Net force on the ring axle arrangement is zero.
∴T1​+T2​=mg

Magnetic torque is
τB​​=i⋅πR2⋅Bsin90∘=iπR2B​
Direction of the torque is as shown in figure.
Torque of T2​ (about O ) is also in same direction. Torque due to T1​ opposes the resultant of torques T2​l and τB​.
∴ or, ​T1​l=T2​l+iπR2BT1​−T2​=li⋅πR2B​​
Solving (i) and (ii) gives
T1​=2mg​+2li⋅πR2B​
and T2​=2mg​−2li⋅πR2B​

Question 10

A current carrying wire is bent in L shape as shown. Find the direction of magnetic field at point P due to the current in the wire.

Accepted Answer

Field at P due to straight wire AB as well as due to BC is directed into the plane of the figure.

Question 11

A long wire, carrying current I, is bent in L shape. Find the magnetic field at point P.

Accepted Answer

Field at P due to the straight segment 1 is zero.
Field at P due to the straight segment 2 (a semiinfinite wire) is
B=4πrμ0​I​⊗

This is the resultant field at P.

Question 12

A square loop of side length a is made of a uniform wire. A current I enters the loop at point A and leaves at C. The wire carrying the current to A and that taking the current out of C are along the diagonal. Find the magnetic field at the centre (O) of the square.

Accepted Answer

Current will divide equally in the parallel paths ABC and ADC. Fields due to each side has same magnitude (at point O ). Direction of field due to AB and BC is into the plane of the figure and that due to AD and DC is out of the plane of the figure. No field is contributed by the wire carrying the current I to A and that going out at C. The point O lies on the extension of these wires.
Therefore, resultant field at O is zero.

Question 13

The figure shows two long straight current carrying wires having current I and 2I. The wires are normal to the plane of the figure and are at separation d. Find a point on the x-axis where the magnetic field is zero.

Accepted Answer

Consider a point P between the two wires. The magnetic field due to the two wires are in opposite directions at P. For the resultant field

to be zero at P, the two fields must be of same magnitude.
⇒​B1​=B2​4πxμ0​I​=4π(d−x)μ0​(2I)​d−x=2x⇒x=3d​​

Question 14

The figure shows two parallel wires having current I in opposite direction. The wires are perpendicular to the plane of the figure and separation between them is 2d. Find the magnetic field produced at point P due to the two wires. Distance MP=d.

Accepted Answer

From the geometry of the figure, ∠APB=90∘. The magnetic field at P due to the wire at A is

BA​=4πrμ0​I​=4π(2​d)μ0​I​
Direction of this field is perpendicular to AP, along PB. Field due to the wire at B is along PA. [To visualise this you may draw circular field lines around the wire].
BB​=BA​=4π(2​d)μ0​I​
The resultant of BA​ and BB​ is along PM and its magnitude is
B=BA2​+BB2​​=4πdμ0​I​

Question 15

An insulated straight wire of length L carries current I. It is bent to form a coil of n turns. Find the magnetic field at the centre of the coil.

Accepted Answer

Let the radius of the coil be r. Then n2πr=L⇒r=2πnL​
Field at the centre is B=n2rμ0​I​
 or, B=n2(2πnL​)μ0​I​=Lμ0​πn2I​

Question 16

A long wire carrying current I is bent into the shape shown in the figure. ACB is 43​ of a circle of radius R and the remaining two arms of the wire are parallel. Find the magnetic field at the centre (O) of the circle.

Accepted Answer

Field at O due to the straight part joining the circle at A is zero. B1​=0
Field at O due to the circular are is
B2​=2Rμ0​I​⋅43​=8R3μ0​I​⊙
Field at O due to the straight part joining the circle at B is
B3​=4πRμ0​I​⊙ [Semi infinite wire]
∴ Resultant field B=4Rμ0​I​(23​+π1​)⊙

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Solutions for Physics for JEE Main and Advanced Magnetism and EMI by Shashi Bhushan Tiwari

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