A proton of mass 1 . 67 × 10 - 27 k g and charge 1 . 6 × 10 - 19 C is projected with a speed of 2 × 10 6 m / s at an angle of 60 ° to the X-axis . If a uniform magnetic field of 0.104 Tesla is applied along Y-axis , the path of the proton is:

A helix of radius = 0.1 m and time period 2 π × 10 - 7 s

A circle of radius = 0.2 m and time period π × 10 - 7 s

A circle of radius = 0.1 m and time period 2 π × 10 - 7 s

A helix of radius = 0.2 m and time period 4 π × 10 - 7 s

A particle of charge +q and mass m moving under the influence of a uniform electric field E i ^ and a uniform magnetic field B k ^ follows trajectory from P to Q as shown in figure. The velocities at P and Q are v i ^ and - 2 v j ^ respectively. Which of the following statement(s) is/are correct?

Rate of work done by electric field at P is 3 4 mv 3 a

Rate of work done by both fields at Q is zero

Physics - Moving Charges Magnetism - Quiz-4

A circular current carrying coil has a radius R. The distance from the centre of the coil on the axis where the magnetic induction will be $\frac{1}{8} th$ to its value at the centre of the coil, is

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$\frac{R}{\sqrt{3}}$
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$R \sqrt{3}$
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$2 \sqrt{3} R$
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$\frac{2}{\sqrt{3}} R$

The sensitivity of a moving coil galvanometer can be increased by decreasing:

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the number of turns in the coil.
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the area of the coil.
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the magnetic field.
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the couple per unit twist of the suspension.

A metallic loop is placed in a magnetic field. If a current is passed through it, then

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The ring will feel a force of attraction
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The ring will feel a force of repulsion
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It will move to and fro about its centre of gravity
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None of these

A straight wire carrying a current i is turned into a circular loop. If the magnitude of the magnetic moment associated with it in M.K.S. unit is M, the length of wire will be

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$4 πiM$
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$\sqrt{\frac{4 πM}{i}}$
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$\sqrt{\frac{4 πi}{M}}$
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$\frac{M π}{4 i}$

A 100 turns coil shown in figure carries a current of 2 amp in a magnetic field $B = 0.2 W b / m^{2}$ . The torque acting on the coil is

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() 0.32 Nm tending to rotate the side AD out of the page
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() 0.32 Nm tending to rotate the side AD into the page
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() 0.0032 Nm tending to rotate the side AD out of the page
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() 0.0032 Nm tending to rotate the side AD into the page

A current of 5 amperes is flowing in a wire of length 1.5 meters. A force of 7.5 N acts on it when it is placed in a uniform magnetic field of 2 Tesla. The angle between the magnetic field and the direction of the current is:

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30 $°$
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45°
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60°
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90°

The field normal to the plane of a coil of n turns and radius r which carries a current i is measured on the axis of the coil at a small distance h from the centre of the coil. This is smaller than the field at the centre by the fraction

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$\frac{3}{2} \frac{h^{2}}{r^{2}}$
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$\frac{2}{3} \frac{h^{2}}{r^{2}}$
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$\frac{3}{2} \frac{r^{2}}{h^{2}}$
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$\frac{2}{3} \frac{r^{2}}{h^{2}}$

In hydrogen atom, the electron is making $6.6 \times 10^{15} r e v / s e c$ around the nucleus in an orbit of radius 0.528 Å. The magnetic moment $(A - m^{2})$ will be

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$1 \times 10^{- 15}$
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$1 \times 10^{- 10}$
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$1 \times 10^{- 23}$
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$1 \times 10^{- 27}$

A proton of mass $1.67 \times 10^{- 27} k g$ and charge $1.6 \times 10^{- 19} C$ is projected with a speed of $2 \times 10^{6} m / s$ at an angle of $60 °$ to the X-axis. If a uniform magnetic field of 0.104 Tesla is applied along Y-axis, the path of the proton is:

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A circle of radius = 0.2 m and time period $π \times 10^{- 7} s$
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A circle of radius = 0.1 m and time period $2 π \times 10^{- 7} s$
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A helix of radius = 0.1 m and time period $2 π \times 10^{- 7} s$
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A helix of radius = 0.2 m and time period $4 π \times 10^{- 7} s$

The magnetic field at the centre of a circular coil of radius r is $π$ times that due to a long straight wire at a distance r from it, for equal currents. Figure here shows three cases : in all cases the circular part has radius r and straight ones are infinitely long. For same current the B field at the centre P in cases 1, 2, 3 have the ratio

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() $(\frac{- π}{2}) : (\frac{π}{2}) : (\frac{3 π}{4} - \frac{1}{2})$
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() $(- \frac{π}{2} + 1) : (\frac{π}{2} + 1) : (\frac{3 π}{4} + \frac{1}{2})$
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() $- \frac{π}{2} : \frac{π}{2} : 3 \frac{π}{4}$
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() $(- \frac{π}{2} - 1) : (\frac{π}{2} - \frac{1}{4}) : (\frac{3 π}{4} + \frac{1}{2})$

Two straight long conductors AOB and COD are perpendicular to each other and carry currents $i_{1}$ and $i_{2}$ . The magnitude of the magnetic induction at a point P at a distance a from the point O in a direction perpendicular to the plane ACBD is:

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$\frac{μ_{0}}{2 πa} (i_{1} + i_{2})$
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$\frac{μ_{0}}{2 πa} (i_{1} - i_{2})$
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$\frac{μ_{0}}{2 πa} {(i_{1}^{2} + i_{2}^{2})}^{1 / 2}$
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$\frac{μ_{0}}{2 πa} \frac{i_{1} i_{2}}{(i_{1} + i_{2})}$

The charge on a particle Y is double the charge on particle X. These two particles X and Y after being accelerated through the same potential difference enter a region of the uniform magnetic field and describe circular paths of radii $R_{1}$ and $R_{2}$ respectively. The ratio of the mass of X to that of Y is:

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${(\frac{2 R_{1}}{R_{2}})}^{2}$
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${(\frac{R_{1}}{2 R_{2}})}^{2}$
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$\frac{R_{1}^{2}}{2 R_{2}^{2}}$
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$\frac{2 R_{1}}{R_{2}}$

In the given figure, the electron enters into the magnetic field. It deflects in ...... direction

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+ ve X direction
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– ve X direction
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+ ve Y direction
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– ve Y direction

If the angular momentum of an electron is $\vec{J}$ then the magnitude of the magnetic moment will be

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$\frac{e J}{m}$
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$\frac{e J}{2 m}$
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ej.2m
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$\frac{2 m}{e J}$

A cell is connected between the points A and C of a circular conductor ABCD of centre O with angle AOC = $60 °$ . If $B_{1}$ and $B_{2}$ are the magnitudes of the magnetic fields at O due to the currents in ABC and ADC respectively, the ratio $\frac{B_{1}}{B_{2}}$ is:

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0.2
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6
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1
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5

An electron, a proton, a deuteron and an alpha particle, each having the same speed are in a region of constant magnetic field perpendicular to the direction of the velocities of the particles. The radius of the circular orbits of these particles are respectively $R_{e}$ , $R_{p}$ , $R_{d}$ and $R_{α}$ . It follows that

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$R_{e} = R_{p}$
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$R_{p} = R_{d}$
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$R_{d} = R_{α}$
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$R_{p} = R_{α}$

An infinitely long conductor PQR is bent to form a right angle as shown. A current I flows through PQR. The magnetic field due to this current at the point M is H₁. Now another infinitely long straight conductor QS is connected at Q so that the current is I/2 in QR as well as in QS, The current in PQ remaining unchanged. The magnetic field at M is now $H_{2}$ The ratio $H_{1} / H_{2}$ is given by

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() $\frac{1}{2}$
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() 1
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() $\frac{2}{3}$
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() 2

A very long straight wire carries a current I. At the instant when a charge +Qat point P has velocity $\vec{v}$ , as shown, the force on the charge is:

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Opposite to OX
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Along OX
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Opposite to OY
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4. Along OY

An electric field of 1500 V / m and a magnetic field of 0.40 weber / $m e t e r^{2}$ act on a moving electron. The minimum uniform speed along a straight line the electron could have is

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(1) $1.6 \times 10^{15} m / s$
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(2) $6 \times 10^{- 16} m / s$
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(3) $3.75 \times 10^{3} m / s$
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(4) $3.75 \times 10^{2} m / s$

A coil having N turns is wound tightly in the form of a spiral with inner and outer radii a and b respectively. When a current I passes through the coil, the magnetic field at the centre is:

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$\frac{μ_{0} NI}{b}$
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$\frac{2 μ_{0} NI}{a}$
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$\frac{μ_{0} NI}{2 (b - a)} \ln \frac{b}{a}$
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$\frac{μ_{0} I^{N}}{2 (b - a)} \ln \frac{b}{a}$

An electron, moving in a uniform magnetic field of induction of intensity $\overset{⇀}{B,}$ has its radius directly proportional to :

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Its charge
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Magnetic field
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Speed
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None of these

The magnetic field due to a straight conductor of a uniform cross-section of radius 'a' carrying steady current is represented by:

A non-planar loop of conducting wire carrying a current I is placed as shown in the figure. Each of the straight sections of the loop is of length 2a. The magnetic field due to this loop at the point P (a,0,a) points in the direction

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$\frac{1}{\sqrt{2}} (- \hat{j} + \hat{k})$
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$\frac{1}{\sqrt{3}} (\hat{i} + \hat{j} + \hat{k})$
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$\frac{1}{\sqrt{3}} (- \hat{j} + \hat{k} + \hat{i})$
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$\frac{1}{\sqrt{2}} (\hat{i} + \hat{k})$

A particle of charge q and mass m is moving along the x-axis with a velocity v and enters a region of electric field E and magnetic field B as shown in the figure below. For which figure the net force on the charge may be zero?

A wire carrying a current i is placed in a uniform magnetic field in the form of the curve $y = α \sin (\frac{πx}{L}), 0 \leq x \leq 2 L$ . The force acting on the wire is:

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$\frac{i B L}{π}$
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$i B L π$
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$2 i B L$
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Zero

A long straight wire along the z-axis carries a current I in the negative z-direction. The magnetic field vector $\vec{B}$ at a point having coordinates (x, y) in the z = 0 plane is :

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$\frac{μ_{0} I (y \hat{i} - x \hat{j})}{2 π (x^{2} + y^{2})}$
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$\frac{μ_{0} I (x \hat{i} + y \hat{j})}{2 π (x^{2} + y^{2})}$
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$\frac{μ_{0} I (x \hat{j} - y \hat{i})}{2 π (x^{2} + y^{2})}$
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$\frac{μ_{0} I (x \hat{i} - y \hat{j})}{2 π (x^{2} + y^{2})}$

A circular coil is in y-z plane with centre at origin. The coil is carrying a constant current. Assuming direction of magnetic field at x = – 25 cm to be positive direction of magnetic field, which of the following graphs shows variation of magnetic field along x-axis

The ratio of the magnetic field at the centre of a current carrying circular wire and the magnetic field at the centre of a square coil made from the same length of wire will be

(a) $\frac{π^{2}}{4 \sqrt{2}}$ (b) $\frac{π^{2}}{8 \sqrt{2}}$

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1
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2
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3
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4

A particle of charge +q and mass m moving under the influence of a uniform electric field $E \hat{i}$ and a uniform magnetic field $B \hat{k}$ follows trajectory from P to Q as shown in figure. The velocities at P and Q are $v \hat{i}$ and $- 2 v \hat{j}$ respectively. Which of the following statement(s) is/are correct?

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$E = \frac{3}{4} \frac{{mv}^{2}}{qa}$
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Rate of work done by electric field at P is $\frac{3}{4} \frac{{mv}^{3}}{a}$
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Rate of work done by both fields at Q is zero
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All of the above

Figure shows a square loop ABCD with edge length a. The resistance of the wire ABC is r

and that of ADC is 2r. The value of magnetic field at the centre of the loop assuming

uniform wire is

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$\frac{\sqrt{2} μ_{0} i}{3 πa} ⊙$
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$\frac{\sqrt{2} μ_{0} i}{3 πa} \otimes$
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$\frac{\sqrt{2} μ_{0} i}{πa} ⊙$
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$\frac{\sqrt{2} μ_{0} i}{πa} \otimes$

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Practice Physics Quiz Questions and Answers

0:0:1

Practice Physics Quiz Questions and Answers

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