Physics - Moving Charges Magnetism - Quiz-3

In the figure, what is the magnetic field at the point O?

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$\frac{μ_{0} I}{4 πr}$
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$\frac{μ_{0} I}{4 πr} + \frac{μ_{0} I}{2 πr}$
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$\frac{μ_{0} I}{4 r} + \frac{μ_{0} I}{4 πr}$
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$\frac{μ_{0} I}{4 r} - \frac{μ_{0} I}{4 πr}$

The magnetic moment of a current (i) carrying circular coil of radius (r) and number of turns (n) varies as :

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$1 / r^{2}$
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$1 / r$
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r
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$r^{2}$

The current is flowing in the south direction along a power line. The direction of the magnetic field above the power line (neglecting earth's field) is :

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South
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East
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North
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West

For the magnetic field to be maximum due to a small element of current carrying conductor at a point, the angle between the element and the line joining the element to the given point must be :

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0°
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90°
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180°
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45°

An electron and a proton enter a magnetic field perpendicularly. Both have the same kinetic energy. Which of the following is true:

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Trajectory of electron is less curved
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Trajectory of proton is less curved
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Both trajectories are equally curved
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Both move on a straight-line path

A proton, a deuteron, and an $α -$ particle having the same kinetic energy are moving in circular trajectories in a constant magnetic field. If $r_{p}, r_{d}$ and $r_{α}$ denote respectively the radii of the trajectories of these particles, then:

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$r_{α} = r_{p} < r_{d}$
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$r_{a} > r_{d} > r_{p}$
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$r_{α} = r_{d} > r_{p}$
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$r_{p} = r_{d} = r_{α}$

An electron and a proton with equal momentum enter perpendicularly into a uniform magnetic field, then :

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The path of proton shall be more curved than that of electron
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The path of proton shall be less curved than that of electron
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Both are equally curved
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Path of both will be a straight line

One proton beam enters a magnetic field of $10^{- 4} T$ normally, Specific charge = $10^{11} C / k g .$ velocity = $10^{7} m / s .$ What is the radius of the circle described by it:

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$0.1 m$
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$1 m$
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$10 m$
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None of these

The maximum kinetic energy of the positive ion of charge q and mass m in the cyclotron of radius $r_o$ in which applied magnetic field is B, is :

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$\frac{q^{2} B r_{0}}{2 m}$
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$\frac{q B^{2} r_{0}}{2 m}$
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$\frac{q^{2} B^{2} r_{0}^{2}}{2 m}$
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$\frac{q B r_{0}}{2 m^{2}}$

An electron of mass m and charge q is traveling with a speed v along a circular path of radius r at right angles to a uniform of the magnetic field B. If the speed of the electron is doubled and the magnetic field is halved, then resulting path would have a radius of:

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$\frac{r}{4}$
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$\frac{r}{2}$
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$2 r$
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$4 r$

Cyclotron cannot be used to accelerate

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Electrons
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Neutrons
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Positive ions
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Both (1) and (2)

Two particles A and B of masses $m_{A}$ and $m_{B}$ respectively and having the same type of charge are moving in a plane. A uniform magnetic field exists perpendicular to this plane. The speeds of the particles are $v_{A}$ and $v_{B}$ respectively, and the trajectories are as shown in the figure. Then

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$m_{A} v_{A} < m_{B} v_{B}$
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$m_{A} v_{A} > m_{B} v_{B}$
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$m_{A} < m_{B} a n d v_{A} < v_{B}$
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$m_{A} = m_{B} a n d v_{A} = v_{B}$

Magnetic field due to a ring having n turns at a distance x on its axis is proportional to (if r = radius of ring) :

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$\frac{r}{(x^{2} + r^{2})}$
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$\frac{r^{2}}{(x^{2} + r^{2})^{3 / 2}}$
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$\frac{{nr}^{2}}{(x^{2} + r^{2})^{3 / 2}}$
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$\frac{n^{2} r^{2}}{(x^{2} + r^{2})^{3 / 2^{}}}$

A and B are two concentric circular conductors of centre O and carrying currents $i_{1}$ and $i_{2}$ as shown in the adjacent figure. If ratio of their radii is 1 : 2 and ratio of the flux densities at O due to A and B is 1 : 3, then the value of $i_{1} / i_{2}$ is

(a) $\frac{1}{6}$ (b) $\frac{1}{4}$

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PQRS is a square loop made of uniform conducting wire the current enters the loop at P and leaves at S. Then the magnetic field will be

(a) Maximum at the centre of the loop
(b) Zero at the centre of the loop
(c) Zero at all points inside the loop
(d) Zero at all points outside of the loop

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An electric current passes through a long straight wire. At a distance 5 cm from the wire, the magnetic field is B. The field at 20 cm from the wire would be :

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$\frac{B}{6}$
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$\frac{B}{4}$
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$\frac{B}{3}$
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$\frac{B}{2}$

The dimension of the magnetic field intensity B is:

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${MLT}^{- 2} A^{- 1}$
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${MT}^{- 2} A^{- 1}$
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${ML}^{2} {TA}^{- 2}$
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$M^{2} {LT}^{- 2} A^{- 1}$

A wire in the form of a square of side ‘a’ carries a current i. Then the magnetic induction at the centre of the square wire is: (Magnetic permeability of free space = $μ_{0}$ )

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$\frac{μ_{0} i}{2 πa}$
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$\frac{μ_{0} i \sqrt{2}}{πa}$
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$\frac{2 \sqrt{2} μ_{0} i}{πa}$
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$\frac{μ_{0} i}{\sqrt{2} πa}$

Four wires each of length 2.0 metres are bent into four loops P, Q, R and S and then suspended into uniform magnetic field. Same current is passed in each loop. Which statement is correct?

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Couple on loop S will be the highest
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Couple on loop P will be the highest
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Couple on loop Q will be the highest
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Couple on loop R will be the highest

A beam of ions with velocity $2 \times 10^{5} m / s$ enters normally into a uniform magnetic field of $4 \times 10^{- 2}$ tesla. If the specific charge of the ion is $5 \times 10^{7}$ C/kg , then the radius of the circular path described will be :

(a) 0.10 m (b) 0.16 m

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If the direction of the initial velocity of the charged particle is perpendicular to the magnetic field, then the orbit will be
or
The path executed by a charged particle whose motion is perpendicular to magnetic field is :

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A straight line
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An ellipse
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A circle
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A helix

A proton and an $α -$ particle enter a uniform magnetic field perpendicularly with the same speed. If proton takes 25 $μ$ sec to make 5 revolutions, then the periodic time for the $α -$ particle would be :

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50 $μ$ sec
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25 $μ$ sec
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10 $μ$ sec
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5 $μ$ sec

An $α -$ particle travels in a circular path of radius 0.45 m in a magnetic field $B = 1.2 W b / m^{2}$ with a speed of $2.6 \times 10^{7} m / s e c$ . The period of revolution of the $α -$ particle is :

(a) $1.1 \times 10^{- 5}$ sec (b) $1.1 \times 10^{- 6}$ sec

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A rectangular loop carrying a current i is situated near a long straight wire such that the wire is parallel to the one of the sides of the loop and is in the plane of the loop. If a steady current I is established in wire as shown in figure, the loop will

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Rotate about an axis parallel to the wire
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Move away from the wire or towards right
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Move towards the wire
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Remain stationary

Two thin long parallel wires separated by a distance b are carrying a current i amp each. The magnitude of the force per unit length exerted by one wire on the other is

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$\frac{μ_{0} i^{2}}{b^{2}}$
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$\frac{μ_{0} i^{2}}{2 πb}$
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$\frac{μ_{0} i}{2 πb}$
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$\frac{μ_{0} i}{2 {πb}^{2}}$

To make the field radial in a moving coil galvanometer :

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The number of turns in the coil is increased
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Magnet is taken in the form of horse-shoe
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Poles are cylindrically cut
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The coil is wounded on the aluminum frame

In a moving coil galvanometer, the deflection of the coil $θ$ is related to the electrical current i by the relation

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$i \propto \tan θ$
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$i \propto θ$
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$i \propto θ^{2}$
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$i \propto \sqrt{θ}$

A moving coil galvanometer has N number of turns in a coil of effective area A, it carries a current I. The magnetic field B is radial. The torque acting on the coil is

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$N A^{2} B^{2} I$
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$N A B I^{2}$
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$N^{2} A B I$
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$N A B I$

If a particle of charge $10^{- 12}$ coulomb moving along the $\hat{x} -$ direction with a velocity $10^{5}$ m/s experiences a force of $10_{}^{- 10}$ newton in $\hat{y}$ - direction due to magnetic field, then the minimum magnetic field is

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$6.25 \times 10^{3}$ tesla in $\hat{z}$ - direction
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$10^{- 15}$ tesla in $\hat{z}$ - direction
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$6.25 \times 10^{- 3}$ tesla in $\hat{z}$ - direction
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$10^{- 3}$ tesla in $\hat{z}$ - direction

A small coil of N turns has an effective area A and carries a current I. It is suspended in a horizontal magnetic field $\dot{B}$ such that its plane is perpendicular to $\dot{B}$ . The work done in rotating it by $180 °$ about the vertical axis is

(a) $N A I B$ (b) $2 N A I B$

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Practice Physics Quiz Questions and Answers

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Practice Physics Quiz Questions and Answers

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