Exercises of Magnetic Fields due to Electric Current - Grad Plus

# 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?
[Ans: 196 T]

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).
[Ans: 2.5 x 10-5T]

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]
[Ans: 18.7 MHz, 2.53 keV T]

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]
[Ans: 6.62 x 10-27 kg]

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]
[Ans: Repulsive, 2.75 x 10-4 kg]

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]
[Ans: 8.35 x 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?
[Ans: 56.7 A]

9. Magnetic field at a distance 2.4 cm from a long straight wire is 16 μT. What must be current through the wire?
[Ans: 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?
[Ans: 5.954 x 10-2 A.m2]

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.
[Ans: 1.49 x 10-5 T, 1.68 x 10-6 T]

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?
[Ans: 3.142 x 10-4 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.
[Ans: 2.34 x 10-8 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°.
[Ans: 1.2 x 10-5 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.
[Ans: 2 x 10-3]

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?
[Ans: 25 A]

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)
[Ans: 7.515 MeV]

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.
[Ans: $latex B\mathit=\frac{\mu_{\mathit0}}{\mathit4\pi}\frac IR\mathit\lbrack\frac{\mathit3\pi}{\mathit2}\mathit+\sqrt{\mathit2}\mathit\rbrack$

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

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