Solutions for Exercises of Magnetic Fields due to Electric Current
1. Choose the correct option.
i) A conductor has 3 segments; two straight and of length L each and a semicircular with radius R. It carries a current I. What is the magnetic field B at point P?
(A) $latex \frac{\mu_{\mathit0}}{\mathit4\pi}\frac IR$
(B) $latex \frac{\mu_{\mathit0}}{\mathit4\pi}\frac I{R^2}$
(C) $latex \frac{\mu_{\mathit0}}{\mathit4}\frac IR$
(D) $latex \frac{\mu_{\mathit0}I}{\mathit4\pi}$
ii) Figure a, b show two Amperian loops associated with the conductors carrying current I in the sense shown.The $latex \oint\overrightarrow B.\overrightarrow{dl}$ in the cases a and b will be, respectively,
(A) $latex -\;\mu_{\mathit0}I,\;0$
(B) $latex \;\mu_{\mathit0}I,\;0$
(C) $latex \;0,\;\mu_{\mathit0}I$
(D) $latex \;0,\;-\;\mu_{\mathit0}I$
iii) A proton enters a perpendicular uniform magnetic field B at origin along the positive x axis with a velocity v as shown in the figure. Then it will follow the following path. [The magnetic field is directed into the paper]
(A) It will continue to move along positive x axis.
(B) It will move along a curved path, bending towards negative y axis.
(C) It will move along a curved path, bending towards positive x axis.
(D) It will move along a sinusoidal path along the positive x axis.
iv) A conducting thick copper rod of length 1 m carries a current of 15 A and is located on the Earth’s equator. There the magnetic flux lines of the Earth’s magnetic field are horizontal, with the field of 1.3 x 10-4 T, south to north. The magnitude and direction of the force on the rod, when it is oriented so that current flows from west to east, are
(A) 14 x 10-4N, downward.
(B) 20 x 10-4N, downward.
(C) 14 x 10-4N, upward.
(D) 20 x 10-4N, upward.
v) A charged particle is in motion having initial velocity $latex \overrightarrow{\mathrm v}$ when it enter into a region of uniform magnetic field perpendicular to $latex \overrightarrow{\mathrm v}$ . Because of the magnetic force the kinetic energy of the particle will
(A) remain uncharged.
(B) get reduced.
(C) increase.
(D) be reduced to zero.
2. A piece of straight wire has mass 20 g and length lm. It is to be levitated using a current of 1 A flowing through it and a perpendicular magnetic field B in a horizontal direction. What must be the magnetic of B?
m = 20 g, l = 1 m, I = 1 A, B = ?
We know that current carrying wire placed in the magnetic field experience force. To levitate this wire, the force must be equal to weight of the wire.
∴ Fmagnetic = mg But, Fmagnetic = IlB
$latex \therefore IlB=mg$
$latex \therefore B=\frac{mg}{Il}=\frac{20\times9.8}{1\times1}=196T$
3. Calculate the value of magnetic field at a distance of 2 cm from a very long straight wire carrying a current of 5 A (Given: μo = 4π x 10-7 Wb/Am).
R = 2 cm = 2 × 10 m, I = 5 A, B =?
For long straight wire,
$latex i.e\;B=\frac{\mu_0I}{2\pi R}$
$latex \therefore\;B=\frac{\mu_0I}{2\pi R}=\frac{4\pi\times10^{-7}\times5}{2\pi\times2\times10^{-2}}$
$latex =\;5\;\times\;10^{-7}\;T$
4. An electron is moving with a speed of 3 x 10-7 m/s in a magnetic field of 6 x 10-4 T perpendicular to its path. What will be the radium of the path? What will be frequency and the energy in keV ? [Given: mass of electron = 9 x10-31 kg, charge e = 6 x 10-19 C, 1 eV = 1.6 x 10-19 J]
v = 3 × 107 m, B = 6 × 10-4 m, r = ?, f = ?, E = ?
$latex qvB\;=\;\frac{mv^2}r\;\;\;\;\;\;\therefore\;r=\frac{mv}{qB}$
$latex \therefore\;r\;=\;\frac{mv}{qB}=\frac{9\times10^{-31}\times3\times10^7}{1.6\times10^{-19}\times6\times10^{-4}}=0.281\;m$
$latex \therefore\;f\;=\;\frac1T\;=\;\frac1{\displaystyle2\pi r/v}$
$latex \therefore\;f\;\;=\;\frac{\displaystyle v}{\displaystyle2\pi r}=\frac{\displaystyle qB}{\displaystyle2nm}$
$latex\therefore\;f\;=\;\frac{1.6\times10^{-1}\times6\times10^{-4}}{2\pi\times9\times10^{-31}}$
$latex \therefore\;f\;=\;0.1868\;\times\;10^8$
$latex \therefore\;f\;=\;18.68\;\times\;10^6\;Hz\;=\;18.68\;MHz\;$
$latex E\;=\;K.E.=\frac12mv^2$
$latex \therefore\;E\;=\frac12\times9\times10^{-31}\times\left(3\times10^7\right)^2$
$latex =40.5\;\times\;10.17\;J$
$latex\therefore\;E=\frac{40.5\;\times\;10.17}{1.6\times10^{-19}}eV$
$latex \therefore\;E\;=\;25.31\;\times\;10^2\;$
$latex \therefore\;E=2.531\times10eV=2.531keV$
5. An alpha particle (the nucleus of helium atom) (with charge +2) is accelerated and moves in a vacuum tube with kinetic energy = 10 MeV. It passes through a uniform magnetic field of 1.88 T, and traces a circular path of radius 24.6 cm. Obtain the mass of the alpha particle. [1 eV = 1.6 x 10-19 J, charge of electron = 1.6 x 10-19 C]
q = 2e = 2 × 1.6 × 10-19C , E = 10MeV
∴ E = 10 × 106 × 1.6 × 10-1 J = 1.6 × 10-1 J
B = 1.88 T, r = 24.6cm = 24.6 × 10-2m, mα = ?
When charge traces circular path in magnetic field,
$latex qvB=\frac{mv^2}r\;\;\;\;\therefore v=\frac{qBr}m$
$latex \therefore\;E\;=\;K.E.=\frac12mv^2=\frac{\left(qBr\right)^2}{2m}$
$latex \therefore\;m\;=\;\frac{\left(qBr\right)^2}{2m}$
$latex \therefore\;m_\alpha\;=\;\frac{\left(q\alpha Br\right)^2}{2E}$
$latex =\;\frac{\left(2\times1.6\times10^{-19}\times1.88\times24.6\times10^{-2}\right)^2}{2\times1.6\times10^{-12}}$
$latex \therefore\;m_\alpha\;=\;6.681\;\times\;10^{-27}\;kg$
Hint : you may use log
6. Two wires shown in the figure are connected in a series circuit and the same amount of current of 10 A passes through both, but in apposite directions. Separation between the two wires is 8 mm. The length AB is S = 22 cm. Obtain the direction and magnitude of the magnetic field due to current in wire 2 on the section AB of wire 1. Also obtain the magnitude and direction of the force on wire 1. [μo= 4π x 10-7 T.m/A]
I1 = 10A, I2 = 10,d = 8mm = 8 × 10-3 m,
S = L = 22cm = 22 × 10-2 m, B =?, F21 =?
Magnetic Filed die to wire -2 at the position of wire -1,
$latex B_{21}\;=\;\frac{\mu_0I_2}{2\pi d}$
$latex =\frac{4\pi\times10^{-7}\times10\times10\times22\times10^{-2}}{2\pi\times8\times10^{-3}}$
a) Magnetic field at the centre of current carrying coil, is given as,
$latex B\;=\;\frac{\mu_0I}{2r}$
$latex =\;\frac{4\pi\times10^{-7}\times2.3}{2\times9.7\times10^{-2}}$
$latex =\;1.489\;\times\;10^{-5}T$
b) Magnetic field at the axial point of current carrying coil, is given as,
$latex \therefore\;B\;=\;\frac{\mu_0Ir^2}{2\left(z^2+r^2\right)^{\displaystyle3/2}}$
$latex =\;\frac{4\pi\times10^{-7}\times2.3\times\left(9.7\times10^{-2}\right)^2}{2\lbrack{(9.7\times10^{-2})}^2+{(9.7\times10^{-2})}^2\rbrack^{\displaystyle3/2}}$
$latex =\;\frac{2\pi\times10^{-7}\times2.3\times\left(9.7\times10^{-2}\right)^2}{2\lbrack{(9.7\times10^{-2})}^2\rbrack^{\displaystyle3/2}}$
$latex \therefore\;B\;=\;1.678\;\times\;10^{-6}T\;\;(Use\;\log)\;$
7. A very long straight wire carries a current 5.2 A. What is the magnitude of the magnetic field at a distance 3.1 cm from the wire? [ μo= 4π x 10-7 T.m/A]
I = 5.2 A, R = 3.1cm = 3.1× 10-2 m, B = ?
For long straight wire,
$latex B\;=\;\frac{\mu_0I}{2\pi R}$
$latex =\;\frac{4\pi\times10^{-7}\times5.2}{2\pi\times3.1\times10^{-2}}$
$latex =\;3.35\;\times\;10^{-5}\;T$
8. Current of equal magnitude flows through two long parallel wires having separation of 1.35 cm. If the force per unit length on each of the wires in 4.76 x 10-2 N, what must be I?
I1 = I2 = I =?, d = 1.35cm = 1.35 × 10-2 m
F/L = f = 4.76 × 10-2N
For two parallel current carrying wire, force per unit length is given as,
$latex f=\frac{\mu_0I_1I_2}{2\pi d}=\frac{\mu_0I^2}{2\pi d}$
$latex \therefore\;I^2\;=\;\frac{2\pi df}{\mu_0}$
$latex \therefore\;I^2\;=\;\frac{2\pi\times1.35\times10^{-2}\times4.76\times10^{-2}}{4\pi\times10^{-7}}$
$latex \therefore\;I^2\;=\;3213$
$latex \therefore\;I\;=\;\sqrt{3213}\;=\;56.7A\;(u\sin g\;\log)\;$
9. Magnetic field at a distance 2.4 cm from a long straight wire is 16 μT. What must be current through the wire?
I =?, R = 2.4 cm = 2.4 × 10-2 m,
B = 16μT = 16 × 10-6T
For long straight wire,
$latex B=\frac{\mu_0I}{2\pi R}$
$latex \therefore\;I\;=\frac{2\pi RB}{\mu_0}$
$latex \therefore\;I\;=\frac{2\pi\times2.4\times10^{-4}\times16\times10^{-6}}{4\pi\times10^{-7}}$
$latex \therefore\;I\;=1.92\;A$
10. The magnetic field at the centre of a circular current carrying loop of radius 12.3 cm is 6.4 x 10-6 T. What will be the magnetic moment of the loop?
r = 12.3 cm = 12.3 × 10-2 m, B = 6.4 ×10-6T
µ = ?
Magnetic field at the centre of current carrying coil, is given as,
$latex B\;=\;\frac{\mu_0I}{2\pi r}$
Magnetic moment of the coil, assuming single turn is given as,
$latex \mu=IA=I\pi r^2$
Put I in the expression of B, we have,
$latex B\;=\;\frac{\mu_0\left({\displaystyle\frac\mu{\pi r^2}}\right)}{2r}$
$latex \therefore\;\mu=\frac{2\pi Br^2}{\mu_0}$
$latex \therefore\;\mu\;=\frac{2\pi\times6.4\times10^{-6}\left(12.3\times10^{-2}\right)^2}{4\pi\times10^{-7}}$
$latex =\;5.954\;\times\;10^{-2}Am^2\;$
11. A circular loop of radius 9.7 cm carries a current 2.3 A. Obtain the magnitude of the magnetic field (a) at the centre of the loop and (b) at a distance of 9.7 cm from the centre of the loop but on the axis.
r = 9.7 cm = 9.7 × 10-2 m, I = 2.3 A,
a) Bcenter = ? (b) Baxis = ? z = 9.7 cm = 9.7 × 10-2 m
a) Magnetic field at the centre of current carrying coil, is given as,
$latex B\;=\;\frac{\mu_0I}{2\pi r}$
$latexB\;=\;\frac{4\times10^{-7}\times2.3}{2\times9.7\times10^{-2}}$
$latex B=\;1.489\times10^{-5}T$
b) Magnetic field at the axial point of current carrying coil, is given as,
$latex B\;=\;\frac{\mu_0Ir^{2^{}}}{2\left(z^2+r^2\right)^{\displaystyle\frac32}}$
$latex B\;=\;\frac{4\pi\times10^{-7}\times2.3\times\left(9.7\times10^{-2}\right)^2}{2\left[\left(9.7\times10^{-2}\right)^2+\left(9.7\times10^{-2}\right)^2\right]^{3/2}}$
$latex B\;=\;\frac{2\pi\times10^{-7}\times2.3\times\left(9.7\times10^{-2}\right)^2}{2\left[\left(9.7\times10^{-2}\right)^2+^2\right]^{\displaystyle3/2}}$
$latex \therefore\;B\;=\;1.678\;\times\;10^{-6}T\;\;(Use\;\log)\;$
12. A circular coil of wire is made up of 100 turns, each of radius 8.0 cm. If a current of 0.40 A passes through it, what be the magnetic field at the centre of the coil?
N = 100, r = 8 cm = 8 × 10-2 I m, = 0.40 A, B =?
Magnetic field at the centre of current carrying coil, is given as,
$latex \therefore\;B\;=\;\frac{\mu_0NI}{2r}$
$latex \therefore\;B\;=\;\frac{4\pi\times10^{-7}\times100\times0.4}{2\times8\times10^{-2}}$
$latex \therefore\;B\;=\;3.142\;\times10^{-4}T$
13. For proton acceleration, a cyclotron is used in which a magnetic field of 1.4 Wb/m2 is applied. Find the time period for reversing the electric field between the two Ds.
q = 1.6 × 10-19C , m = 1.67 × 10-19kg
B = 1.4 wb/m2, T/2=?
For cyclotron covering both dees we have,
$latex T=\frac{2\pi m}{qB}$
$latex \therefore\;T\;=\;\frac{2\times3.14\times1.67\times10^{-27}}{1.6\times10^{-19}\times1.4}$
$latex \therefore\;T\;=\;4.68\times10^{-9}s$
But as we are asked time of reversing the field,
$latex\therefore\;\frac{T\;}2=\frac{4.68\times10^{-9}}2$
$latex =\;2.34\;\times\;10^{-9}s$
14. A moving coil galvanometer has been fitted with a rectangular coil having 50 turns and dimensions 5 cm x 3 cm. The radial magnetic field in which the coil is suspended is of 0.05 Wb/m2. The torsional constant of the spring is 1.5 x 10-9 Nm/degree. Obtain the current required to be passed through the galvanometer so as to produce a deflection of 30°.
N = 50, A = 5 cm × 3cm = 15cm = 15 ×10-4 m2
$latex B\;=\;0.05\;\frac{wb}{m^2},$
$latex k=1.5\times10^{-19}\frac{Nm}{degree}$
$latex \theta=30^\circ,\;I=?$
For moving coil galvanometer, we have,
$latex NIAB\;=\;K\theta$
$latex \therefore\;I\;=\frac{k\theta}{NAB}$
$latex \therefore\;I\;=\frac{1.5\times10^{-19}\times30}{50\times15\times10^{-4}\times0.05}$
$latex \therefore\;I\;=1.2\times10^{-5}A$
15. A solenoid of length π m and 5 cm in diameter has winding of 1000 turns and carries a current of 5 A. Calculate the magnetic field at its centre along the axis.
$latex L\;=\;\pi m,\;D\;=\;5\;cm\;=\;5\;\times\;10^{-2}\;m,\;I\;=\;5\;A$
$latex N\;=\;1000,\;B\;=?$
Magnetic field inside the solenoid,
$latex B=\;\mu_0NI$
Here n is the number of turns per unit length,
$latex \therefore\;B=\mu_0\frac NLI$
$latex \therefore\;B=\frac{4\pi\times10^{-7}\times1000\times5}\pi$
$latex \therefore\;B=2\times10^{-3}T$
16. A toroid of narrow radius of 10 cm has 1000 turns of wire. For a magnetic field of 5 x 10-2 T along its axis, how much current is required to be passed through the wire?
r = 10 cm = 10-1 m, N = 1000, B= 5×10-2 T, I=?
Magnetic field inside the toroid,
$latex B=\mu_0NI$
Here n is the number of turns per unit length
$latex \therefore\;B\;=\;\mu\frac n{2\pi r}I$
$latex \therefore I=\frac{2\pi rB}{\mu_0N}$
$latex \therefore\;I\;=\frac{2\pi\times10^{-1}\times5\times10^{-2}}{4\pi\times10^{-7}\times1000}$
$latex \therefore\;I\;=\;25\$
17. In a cyclotron protons are to be Radius of its D is 60 cm. and its oscillator frequency is 10 MHz. What will be the kinetic energy of the proton thus accelerated?
(Proton mass = 1.67 x 10-27 kg, e= 1.60x 10-19 C, eV = 1.6x 10-19 J)
R = 60 cm = 60 × 10-2 m, q=1.6×10-1 C,
m = 1.67 × 10-27kg, f = 10MHz=60×106 Hz,
K.E. = ?
Maximum kinetic energy of the charge accelerated by cyclotron is,
$latex \therefore K.E_{max}=\frac{\left[qBr\right]^2}{2m}$
So let us find B first. For cyclotron we have,
$latex f\;=\;\frac{qB}{2\pi m}$
$latex \therefore\;B\;=\;\frac{2\pi mf}q$
$latex \therefore\;B\;=\;\frac{2\pi\times1.67\times10^{-2}\times10\times10^6}{1.6\times10^{-19}}$
$latex \therefore\;B\;=\;0.656T$
$latex\therefore\;K.E_{max}\;=\frac{\;\left[1.6\times10^{-1}\times0.656\times60\times10^{-2}\right]^2}{2\times1.67\times10^{-27}}$
$latex \therefore\;K.E_{max}\;=\;1.18\times10^{12}$
$latex \therefore\;K.E_{max}\;=1.18\times10^{12}J$
$latex =\;\frac{1.18\times10^2}{1.6\times10^{-19}$
$latex \therefore\;K.E_{max}\;=7.516\times10^6eV$
$latex \therefore\;K.E_{max}=7.516MeV$
18. A wire loop of the form shown in the figure carries a current I. Obtain the magnitude and direction of the magnetic field at P.
Magnetic field at P is made of two segments of wire i.e. curved arc and straight part.Hence magnetic field at B is vector addition to them.
$latex {\overrightarrow B}_P={\overrightarrow B}_{arc}+{\overrightarrow B}_{straight}$
$latex For\;{{\overrightarrow B}_{arc},\;}magnetic\;field\;is\;given\;as,$
$latex {\overrightarrow B}_{arc}=\frac{\mu_0I}{4\pi R}\times\varnothing(in\;radian)$
$latex \therefore\;{\overrightarrow B}_{arc}=\frac{\mu_0I}{4\pi R}\times270\times\frac\pi{180}$
$latex \therefore\;{\overrightarrow B}_{arc}=\frac{\mu_0I}{4\pi R}\times\frac{3\pi}2$
$latex According\;to\;right\;hand\;thumb\;rule,\;{\overrightarrow B}_{arc}\;is\;pointing\;into\;the\;paper.\;$
$latex For\;{\overrightarrow B}_{straight},\;magnetic\;field\;is\;given\;as,\;$
$latex {\overrightarrow B}_{straight}=\frac{\mu_0I}{2\pi R}\int_0^\theta\cos\theta d\theta$
Here ϕ is the semi vertical angle measured from the perpendicular distance of the point from the wire
$latex i.e.\;for\;this\;question\;45^\circ\;or\;\frac\pi4$
$latex {\overrightarrow B}_{straight}=\frac{\mu_0I}{2\pi R}\int_0^\frac\pi4\cos\theta d\theta=\frac{\mu_0I}{2\pi R}\sin\phi_0^\frac\pi4$
$latex {\overrightarrow B}_{straight}=\frac{\mu_0I}{2\pi R}\times\frac1{\sqrt2}$
According to right hand thumb rule, Bstraight is pointing into the paper
$latex \therefore{\overrightarrow B}_P={\overrightarrow B}_{arc}+{\overrightarrow B}_{straight}=\frac{\mu_0I}{4\pi R}\times\frac{3\pi}2+\frac{\mu_0I}{2\pi R}\times\frac1{\sqrt2}$
$latex \therefore{\overrightarrow B}_p=\frac{\mu_0I}{4\pi R}\left[\frac{3\pi}2+2\frac1{\sqrt2}\right]$
$latex \therefore{\overrightarrow B}_p=\frac{\mu_0I}{4\pi R}\left[\frac{3\pi}2+\sqrt2\right]$
19. Two long parallel wires going into the plane of the paper are separated by a distance R, and carry a current I each in the same direction. Show that the magnitude of the magnetic field at a point P equidistant from the wires and subtending angle 0 from the plane containing the wires, is $latex B\mathit=\frac{\mu_{\mathit0}}\pi\frac IR\mathit{sin}\mathit2\theta$ What is the direction of the magnetic
field?
20. Figure shows a cylindrical wire of diameter a, carrying a current I The current density which is in the direction of the central axis of the wire varies linearly with radial distance r from the axis according to the relation J= Jo r/a. Obtain the magnetic field B inside the wire at a distance r from its centre.
[Ans: $latex B\mathit=\frac{{\mathit J}_{\mathit0}{\mathit\mu}_{\mathit0}\mathit r^{\mathit2}}{\mathit3\mathit a}$ ]
21. In the above problem, what will be the magnetic field B inside the wire at a distance r from its centre, if the current density J is uniform across the cross section of the wire?[Ans: $latex B\mathit=\frac{\mu_{\mathit0}Jr}\pi$ ]