CBSE Class 12 Physics Magnetic Effects of Electric Current Advanced Problems

CBSE Class 12 Physics Magnetic Effects of Electric Current Advanced Problems. Please refer to the examination notes which you can use for preparing and revising for exams. These notes will help you to revise the concepts quickly and get good marks.

Class 12 Physics: Magnetic Effects of Electric Current (Advanced Problems)

An \(\alpha\)-particle is describing a circle of radius 0.45 metre in a field of magnetic induction of 1.2 weber/metre². The mass of the \(\alpha\)-particle is \(6.8\times10^{-27}\) kg and its charge is twice the charge of an electron, i.e. \(3.2\times10^{-19}\) coulomb.

1. Find its speed, frequency of rotation and kinetic energy.
2. What potential difference is required to accelerate the particle to give this much energy to it?

Answer: (a) \( v=\dfrac{qBr}{m}\approx2.5\times10^7\ \text{m/s} \); \( f=\dfrac{qB}{2\pi m}\approx9.0\times10^6\ \text{Hz} \); \( K=\tfrac12mv^2\approx2.2\times10^{-12}\ \text{J}\approx14\ \text{MeV} \).

(b) \( V=\dfrac Kq\approx6.9\times10^6\ \text{V} \) (about 7 MV).

Note: the printed key gives \(2.6\times10^7\) m/s and \(9.2\times10^6\) Hz, which are slightly high; the data give 2.54×10⁷ m/s and 9.0×10⁶ Hz. The key gives no answer for part (b).

A charge of 1 coulomb is placed at one end of a non-conducting rod of length 0.6 m. The rod is rotated in a vertical plane about a horizontal axis passing through the other end of the rod with angular frequency \(10^4\pi\) radian/sec. Find the magnetic field at a point on the axis of rotation at a distance of 0.8 m from the centre of the path.

Answer: The rotating charge is equivalent to a current \( I=qf=1\times\dfrac{10^4\pi}{2\pi}=5000 \) A in a ring of radius \(R=0.6\) m. On the axis,

\[ B=\frac{\mu_0IR^2}{2(R^2+x^2)^{3/2}}=\frac{4\pi\times10^{-7}\times5000\times0.36}{2\times(1.0)^{3/2}}=1.13\times10^{-3}\ \text{T} \]

Note: the printed key lists two values, \(1.13\times10^{-3}\) T and \(2.26\times10^{-3}\) T. The second would correspond to a frequency of \(10^4\) Hz, which does not match the data given, so the second value, \(2.26\times10^{-3}\) T, is quoted here only as it appears in the printed key.

An \(\alpha\)-particle is accelerated by a potential difference of \(10^4\) V. Find the change in its direction of motion, if it enters normally in a region of thickness 0.1 m having transverse magnetic induction of 0.1 tesla. (mass of \(\alpha\)-particle \(=6.4\times10^{-27}\) kg).

Answer: The radius of the path is \( r=\dfrac{\sqrt{2mqV}}{qB}=0.2 \) m, and \( \sin\theta=\dfrac{0.1}{0.2}=\dfrac12 \), so \( \theta=30^\circ \).

There is a constant homogeneous electric field of 100 V/m within the region \(x=0\) and \(x=0.167\) m pointing along the \(x\) direction. There is a constant homogeneous magnetic field \(B\) within the region \(x=0.167\) m and \(x=0.334\) m pointing along the \(z\)-direction. A proton at rest at the origin (\(x=0,\ y=0\)) is released. Find the minimum strength of the magnetic field \(B\), so that the proton is detected back at \(x=0,\ y=0.167\) m. (mass of the proton \(=1.67\times10^{-27}\) kg)

Answer: \( B=7.07\times10^{-3}\ \text{T} \)

The proton gains kinetic energy \(eEd=2.67\times10^{-18}\) J, so \(v=5.66\times10^4\) m/s at \(x=0.167\) m. In the magnetic region it moves on a semicircle and comes back to \(x=0.167\) m at \(y=2r\), moving in \(-x\). The electric field then stops it exactly at \(x=0\) (by symmetry). So \(2r=0.167\) m, \(r=0.0835\) m, and \( B=\dfrac{mv}{er}=7.07\times10^{-3} \) T.

Note: the source gives the magnetic region as "\(x=1.67\) m and \(x=0.334\) m"; this is a typing error for \(x=0.167\) m to \(0.334\) m.

An electron gun G emits electrons of energy 2 keV travelling in the positive \(x\)-direction. The electrons are required to hit the spot S, where GS \(=0.1\) m, and the line GS makes an angle of 60° with the \(x\)-axis as shown in the figure. A uniform magnetic field \(B\) parallel to GS exists in the region outside the electron gun. Find the minimum value of \(B\) needed to make the electron hit S.

Answer: \( B=4.7\times10^{-3}\ \text{T} \)

The velocity component along \(B\) is \(v\cos60^\circ\) and the electron goes in a helix. It reaches the line GS again after one turn, so \( v\cos60^\circ\times\dfrac{2\pi m}{eB}=0.1 \) with \(v=2.65\times10^7\) m/s (from 2 keV), giving \(B=4.7\times10^{-3}\) T.

A particle of mass \(1\times10^{-26}\) kg and charge \(+1.6\times10^{-19}\) C travelling with a velocity \(1.28\times10^6\) m/s in the \(+x\) direction enters a region in which a uniform electric field \(E\) and a uniform magnetic field \(B\) are present such that \(E_x=E_y=0\), \(E_z=-102.4\) kV/m and \(B_x=B_z=0\), \(B_y=8\times10^{-2}\) Wb/m². The particle enters the region at the origin at time \(t=0\). Determine the location (\(x\), \(y\) and \(z\)-coordinates) of the particle at \(t=5\times10^{-6}\) s. If the electric field is switched off at this instant (with the magnetic field still present), what will be the position of the particle at \(t=7.45\times10^{-6}\) s?

Answer: At \(t=5\times10^{-6}\) s: \( (6.4,\ 0,\ 0) \) m. At \(t=7.45\times10^{-6}\) s: \( (6.4,\ 0,\ 2) \) m.

The electric force \(qE=1.64\times10^{-14}\) N (along \(-z\)) balances the magnetic force \(qvB=1.64\times10^{-14}\) N (along \(+z\)), so the particle moves straight: \(x=vt=6.4\) m. With E off, it moves in a circle of radius \( r=\dfrac{mv}{qB}=1 \) m (centre at \(z=1\) m) with period \( T=\dfrac{2\pi m}{qB}=4.9\times10^{-6} \) s. The extra time \(2.45\times10^{-6}\) s is half a period, so the particle is at the opposite end of the diameter: \(x=6.4\) m, \(z=2\) m.

Two long straight parallel wires are 2 m apart, perpendicular to the plane of the paper (see figure). The wire A carries a current of 9.6 A, directed into the plane of the paper. The wire B carries a current such that the magnetic field of induction at the point P, at a distance of 10/11 m from the wire B, is zero. Find:

1. The magnitude and direction of the current in B.
2. The magnitude of the magnetic field of induction at the point S.
3. The force per unit length on the wire B.

Answer: (a) 3 A, out of the plane of the paper. (b) \(1.3\times10^{-6}\) T. (c) \(2.88\times10^{-6}\) N/m, towards P (the currents are opposite, so the wires repel).

Two circular coils each of 100 turns are held such that one lies in the vertical plane and the other in the horizontal plane with their centres coinciding. The radii of the vertical and horizontal coils are respectively 20 cm and 30 cm. If the directions of the currents in them are such that the earth's magnetic field at the centre of the coils is exactly neutralised, calculate the currents in each coil [\(H=27.8\) A/m, angle of dip \(=30^\circ\)].

Answer: \( I_V=0.111 \) A (vertical coil), \( I_H=0.0963 \) A (horizontal coil).

Two circular rings each of radius \(a\) are joined together such that their planes are perpendicular to each other as shown in the figure. The resistance of each half part of the ring is indicated. A very small loop of mass \(m\) and radius \(r\) carrying a current \(I_0\) is placed in the plane of the paper at the common centre of the rings. The loop can freely rotate about any of its diametric axes. If the loop is slightly displaced, find the time period of its oscillations.

Answer: The four half-rings (1, 2, 1, 1 Ω) are in parallel across 2 V, so they carry 2 A, 1 A, 2 A and 2 A. The two halves of the horizontal ring carry equal currents, so their fields cancel at the centre. The vertical ring gives \( B=\dfrac{\mu_0(2-1)}{4a}=\dfrac{\mu_0}{4a} \). For a small displacement, \( I\ddot\theta=-(I_0\pi r^2)B\,\theta \) with \( I=\tfrac12mr^2 \), so

\[ T=\sqrt{\frac{8\pi^2ma}{\mu_0I_0}} \]

Note: the time period is as given in the printed key.

Consider a non-conducting plate of radius \(r\) and mass \(m\) which has a charge \(q\) distributed uniformly over it. The plate is rotated about its axis with an angular speed \(\omega\). Show that the magnetic moment \(\mu\) and the angular momentum \(l\) of the plate are related as \( \mu=-\dfrac{q}{2m}\,l \).

Answer: Take a ring of radius \(x\) and width \(dx\). Its charge is \( dq=\dfrac{q}{\pi r^2}2\pi x\,dx \), its current is \( dI=\dfrac{dq\,\omega}{2\pi} \), and its moment is \( d\mu=\pi x^2dI=\dfrac{q\omega}{r^2}x^3dx \). So \( \mu=\dfrac{q\omega}{r^2}\int_0^rx^3dx=\dfrac{q\omega r^2}{4} \). The angular momentum is \( l=\tfrac12mr^2\omega \). Therefore

\[ \frac{\mu}{l}=\frac{q}{2m}\ \Rightarrow\ \mu=\frac{q}{2m}\,l \]

Note: the two vectors point the same way for a positive charge; the minus sign in the printed question holds only when \(q\) stands for the magnitude of a negative charge (e.g. electrons).

A current \(i\) flows in a long straight wire with cross-section having the form of a thin half ring of radius \(R\) as shown in figure. Find the induction of magnetic field at the point O.

Answer: \[ B=\frac{\mu_0 i}{\pi^2R} \]

A straight segment OC (of length \(L\) metre) of a circuit carrying a current \(I\) amp is placed along the X-axis. Two infinitely long straight wires A and B, each extending from \(Z=-\infty\) to \(+\infty\), are fixed at \(y=-a\) metre and \(y=+a\) metre respectively as shown in the figure. If the wires A and B each carry a current \(I\) amp into the plane of the paper, obtain the expression for the force acting on the segment OC. What will be the force on OC if the current in the wire B is reversed.

Answer: \[ \vec F=\frac{\mu_0I^2}{2\pi}\ln\left[\frac{L^2+a^2}{a^2}\right](-\hat k) \]

If the current in B is reversed, the net field on the x-axis becomes parallel to OC, so the force on OC is zero.

A long straight wire is coplanar with a current carrying circular loop of radius \(R\) as shown in figure. Current flowing through the wire and the loop is \(I_0\) and \(I\) respectively. If the distance between the centre of the loop and the wire is \(r=2R\), calculate the force of attraction between the wire and the loop.

Answer: \[ F=\mu_0I_0I\left(\frac{2}{\sqrt3}-1\right) \]

Figure shows a wooden cylinder of mass \(m=0.25\) kg, radius \(R\) and length \(l=0.1\) m with \(N=10\) turns of wire wrapped around it longitudinally, so that the plane of the coil contains the axis of the cylinder. What is the least current through the loop that will prevent the cylinder from rolling down a plane whose surface is inclined at an angle \(\theta\) to the horizontal, in the presence of a vertical magnetic field of 0.5 Wb m⁻², if the plane of windings is parallel to the inclined plane?

Answer: \( I=2.45\ \text{A} \)

About the point of contact, the weight gives torque \( mgR\sin\theta \). The coil (area \(2Rl\)) has moment \( NI(2Rl) \) perpendicular to the incline, making angle \(\theta\) with the vertical field, so the magnetic torque is \( NI(2Rl)B\sin\theta \). Balancing, \( I=\dfrac{mg}{2NlB}=\dfrac{0.25\times9.8}{2\times10\times0.1\times0.5}=2.45 \) A.

Note: the printed key says 4.9 A, which is what \( mg/(NlB) \) gives, i.e. it uses a coil width of \(R\) instead of \(2R\). The figure in the source also shows only a rectangular loop abcd (10 cm and 5 cm sides, 30° tilt), not the cylinder.

A hypothetical magnetic field existing in a region is given by \( \vec B=B_0\,\hat e_r \), where \( \hat e_r \) denotes the unit vector along the radial direction. A circular loop of radius \(a\), carrying a current \(i\), is placed with its plane parallel to the X-Y plane and the centre at \((0,0,d)\). Find the magnitude of the magnetic force acting on the loop.

Answer: \[ F=\frac{2\pi a^2\,iB_0}{\sqrt{a^2+d^2}} \]

Electrons are observed to be ejected in various directions with negligible speed from a negative plate of a parallel plate capacitor when the plate is illuminated by light of a certain wavelength (photoelectric effect). The plates are separated by a distance \(d\) and a potential difference \(V\) is maintained between them. Show that none of these electrons will reach the positive plate if a magnetic field is applied at right angles to the electric field and the magnetic induction has a value \( B>\left(\dfrac{2mV}{ed^2}\right)^{1/2} \), where \(m\) and \(e\) are the electron mass and charge respectively.

Answer: Let \(E=V/d\) and let \(y\) be the distance moved along the field. Energy: \( \tfrac12mv^2=eEy \). Momentum along the plate: \( mv_x=eBy \). At the farthest point the electron moves parallel to the plates, so \(v=v_x=eBy/m\); then \( \tfrac12m\left(\dfrac{eBy}m\right)^2=eEy \) gives \( y_{max}=\dfrac{2mE}{eB^2}=\dfrac{2mV}{edB^2} \). No electron reaches the positive plate if \(y_{max}<d\), i.e.

\[ B>\left(\frac{2mV}{ed^2}\right)^{1/2} \]

A beam of charged particles, having kinetic energy \(10^3\) eV, contains masses \(8\times10^{-27}\) kg and \(1.6\times10^{-26}\) kg emerge from the end of an acceleration tube. There is a plate at a distance \(10^{-2}\) m from the end of the tube and placed perpendicular to the beam. Calculate the magnitude of the smallest magnetic field which can prevent the beam from striking the plate.

Answer: \( B=\sqrt2\approx1.4 \) tesla

The radius of the circular path must not exceed \(10^{-2}\) m. The heavier particle has the larger radius, so it fixes the field: \( B=\dfrac{\sqrt{2mK}}{qd}=\dfrac{\sqrt{2\times1.6\times10^{-26}\times1.6\times10^{-16}}}{1.6\times10^{-19}\times10^{-2}}=\sqrt2 \) T (taking the charge of each particle as \(e\), which the question does not state).

A metal wire of mass \(m\) slides without friction on two rails spaced at a distance \(L\) apart. The track lies in a vertical uniform field of induction \(B\). A constant current \(I\) flows along one rail across the wire and back down the other rail. Find the velocity (speed and direction) of the wire as a function of time assuming it to be at rest initially.

Answer: The force on the wire is \(F=BIL\), constant, so \( v=\dfrac{BIL}{m}\,t \), directed along the rails to the left (as in the printed answer).

Two parallel rails separated by a distance of \(2\times10^{-2}\) metre are laid in the north-south direction. On these rails is kept a metal cylinder of mass \(4.0\times10^{-2}\) kg. A battery is connected to the rails with its positive terminal joined to the rail on the eastern side. The battery sends a current of 3.0 amp to the cylinder. If a uniform magnetic field of 1.2 tesla directed upward be there, then find the magnitude and direction of the magnetic force imposed on the cylinder. What would be the acceleration produced in the cylinder? Friction is negligible.

Answer: \( F=BIL=1.2\times3.0\times2\times10^{-2}=7.2\times10^{-2} \) N, directed towards the north (the current flows from the east rail to the west rail through the cylinder, \(\vec F=I\vec L\times\vec B\)). Acceleration \( a=F/m=1.8\ \text{m/s}^2 \) towards the north.

Note: the printed key gives only the magnitudes (7.2×10⁻² N, 1.8 m/s²); the direction is worked out here.