LinkedIn Insight Solutions for Exercises of Magnetic Fields due to Electric Current - Grad Plus

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. o= 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 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 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: $latex B\mathit=\frac{\mu_{\mathit0}Jr}\pi$ ]


 

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