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Physics M.C.Q (1 Marks)

Moving Charges and Magnetism · Rajasthan Board (RBSE) STD 12 Science (Rajasthan - English Medium). 228 questions.

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M.C.Q (1 Marks)

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Question 11 Mark
Consider a wire carrying a steady current, I placed in a uniform magnetic field B perpendicular to its length. Consider the charges inside the wire. It is known that magnetic forces do no work. This implies that:
Answer
  1. Some charges inside the wire move to the surface as a result of B.​​​​​​
  1. If the wire moves under the influence of B, no work is done by the magnetic force on the ions, assumed fixed within the wire.
Solution:
Magnetic force on a conductor of length l carrying a current I placed in a uniform magnetic field B is given by, $\text{F}=\text{I}(\text{l}\times\text{B})\text{ or }|\text{F}|=\text{I }|\text{l}| \ \ |\text{B}|\sin\theta$.
The direction of force is given by Fleming's left hand rule & F is perpendicular to the direction of magnetic field B, Therefore, work done by the magnetic force on the ions is zero.
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Question 21 Mark
A charged particle moves through a magnetic field perpendicular to its direction. Then:
Answer
  1. The momentum changes, but the kinetic energy is constant.
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Question 31 Mark
A rod $AB$ moves with a uniform velocity $v$ in a uniform magnetic field as shown in figure.
Answer
The end $A$ becomes positively charged.
The end $'A\ '$ becomes, positively charged.
Because magnetic field exerts an average Force $\overrightarrow{\text{F}}_0=\text{q}\vec{\text{v}}\times\vec{\text{B} } \ n$ each free electron where $q = 1. 6 \times 10^{19}C$ is the charge on the electron.
This Force is towards $AB$ and hence the free electrons will move towareds $B$.
Negative charge is accumulated at $'B\ '$ and positive charge appears at $A$.
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Question 41 Mark
Beta particle An unknown particle is being studied in a magnetic field of variable intensity and direction. When the magnetic field is turned off, the particle is observed to move toward the earth. When the magnetic field is turned on, the particle is observed to continue to move toward the earth, no matter the strength or the direction of the magnetic field. Which of the particles listed below is most likely the unknown particle?
Answer
  1. Gamma ray
Explanation:
We know that a  moving charged particle experiences a force in a magnetic force , as the given unknown particle is not affected by magnetic field , so it cannot be a charged particle.
But beta, alpha and positron are charged particle therefore they are not the unknown particle , only neutron is chargeless particle so it is the unknown particle.
Gamma rays are not particle though they are also not affected by a magnetic field.
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Question 61 Mark
According to oersted, around a current carrying conductor, magnetic field exists:
Answer
  1. As long as there is current in the wire.
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Question 71 Mark
The path of an electron in a uniform magnetic field may be:
Answer
  1. Either helical or circular
Explanation:
The path an electron in a uniform magnetic field my be either helical or circular.
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Question 101 Mark
A particle of mass m and charge q enters a magnetic field B perpendicularly with a velocity v. The radius of the circular path described by it will be:
Answer
  1. mv/Bq.
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Question 111 Mark
An electron having a charge e moves with a velocity v in X-direction. An electric field acts on it in Y-direction? The force on the electron acts in:
Answer
  1. Negative direction of Y-axis.
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Question 121 Mark
When a charged particle moves through a magnetic field, the quantity which is not affected in the magnetic field is:
Answer
  1. Kinetic energy of the particle
Explanation:
Magnetic field does zero work on moving charged particles. Therefore, kinetic energy of particle remains constant.
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Question 131 Mark
Two parallel, long wires carry currents $i_1$ and $i_2$ with $i_1  >  i_2. $When the currents are in the same direction, the magnetic field at a point midway between the wires is $10 \mu\text{T}.$ If the direction of $i_2$ is reversed, the field becomes $30 \mu\text{T}.$ The ratio $\frac{\text{i}_1}{\text{i}_2}$ is:
Answer
The magnetic field due to the current$-$carrying long, straight wire at point a is given by,$\text{B}=\frac{\mu_0\text{i}}{2\pi\text{d}}$
When both the wires carry currents $i_1$ and $i_2$ ​in the same direction, they produce magnetic fields in opposite directions at any point in between the wires.
$\text{B}'=\frac{\mu_0\text{i}_1}{2\pi\text{a}}-\frac{\mu_0\text{i}_2}{2\pi\text{a}}=10\mu\text{T}\ ...(1)$
Here, a is the distance of the midpoint from both the wires.
When both the wires carry currents in opposite directions, they produce fields in the same direction at the midpoint of the two wires.
$\text{B}''=\frac{\mu_0\text{i}_1}{2\pi\text{a}}+\frac{\mu_0\text{i}_2}{2\pi\text{a}}=30\mu\text{T}\ ...(2)$
On solving eqs. $(1)$ and $(2)$, we get
$\text{i}_1-\text{i}_2=10$
$\text{i}_1+\text{i}_2=30$
$\Rightarrow\text{i}_1=20$
$ \text{i}_2=10$
$\frac{\text{i}_1}{\text{i}_2}=\frac{20}{10}$
$\frac{\text{i}_1}{\text{1}_2}=\frac{2}{1}$
$\Rightarrow\frac{\text{i}_1}{\text{i}_2}=2$
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Question 141 Mark
A moving charge produces:
Answer
  1. Both of them.
Explanation:
Because of the presence of a charge, a particle produces an electric field. Also, because of its motion, that is, the flow of charge or current, there is generation of a magnetic field.
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Question 151 Mark
A positively charged particle projected towards east is deflected towards north by a magnetic field. The field may be:
Answer
  1. Towards west.
Explanation:
$\text{F}=\text{q}\big(\overrightarrow{\text{V}}\times\overrightarrow{\text{B}}\big)$

$\text{j}=\text{q}\big(\overrightarrow{\text{i}}\times\overrightarrow{\text{B}}\big)$
$\Rightarrow\text{B}\otimes$
⇒ The magnetic field may be down ward direction.
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Question 161 Mark
Let $\overrightarrow{\text{E}}$ and $\overrightarrow{\text{B}}$ denote electric and magnetic fields in a frame S and $\overrightarrow{\text{E}}$ and $\overrightarrow{\text{B}}$ in another frame S moving with respect to S at a velocity $\overrightarrow{\text{v}}.$ Two of the following equations are wrong. Identify them.
Answer
  1. $\text{E}_\text{y},=\text{E}_\text{y}+\frac{\text{vB}_\text{z}}{\text{c}^2}$
  2. $\text{B}'_\text{y}=\text{B}_\text{y}+\text{v}\text{E}_\text{z}$
Explanation:
$\text{qE}=\text{qvB}$
$\Rightarrow\text{e}=\text{vB}$ By dimensionally b & care wrong
$\Rightarrow\text{v}\text{E}=\text{v}^2\text{B}$
$\Rightarrow\text{B}=\frac{\text{vE}}{\text{v} ^2}$
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Question 171 Mark
If the direction of the initial velocity of a charged particle is neither along nor perpendicular to that of the magnetic field, then the orbit will be:
Answer
  1. A helix.
Explanation:
The perpendicular component will be responsible for circular motion and parallel component will take it along the magnetic field.
Considering both the phenomena, the resultant motion will be helical.
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Question 181 Mark
What is the space around a current-carrying conductor, in which its magnetic effect can be experienced called?
Answer
  1. Magnetic field
Explanation:
The space around a current-carrying conductor, in which its magnetic effect can be experienced is called the magnetic field.
When a current is passed through a conductor, it modifies the space around the conductor and forms a magnetic field.
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Question 191 Mark
The magnetic field produced by a current - carrying wire at a given point depends on:
Answer
  1. The current passing through it
Explanation:
The magnetic field produced by a current - carrying wire at a given point depends directly on the current passing through it.
Therefore, if there is a circular coil having n turns, the field produced is n times as large as that produced by a single turn.
This is because the current in each circular turn has the same direction, and the field due to each turn then just adds up.
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Question 201 Mark
A current carrying circular loop of radius R is placed in the x-y plane with centre at the origin. Half of the loop with x > 0 is now bent so that it now lies in the y-z plane.
Answer
  1. The magnitude of magnetic moment now diminishes.
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Question 211 Mark
When we double the radius of a coil keeping the current through it unchanged, what happens to the magnetic field directed along its axis at far off points?
Answer
  1. Remains unchanged
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Question 231 Mark
State the rule that is used to find the direction of field acting at a point near a current-carrying straight conductor.
Answer
  1. The right-hand thumb rule
Explanation:
Right-hand thumb rule can be used to find the direction of the magnetic field at a point near a current-carrying conductor.
Right hand rule states that, if the thumb of the right hand is in the direction of the current flow then, the curl fingers show the direction of the magnetic field.
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Question 251 Mark
Magnetic field at the centre of a circular coil of radius r, through which a current I flows is:
Answer
  1. Directly proportional to I.
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Question 261 Mark
A charged particle moves along a circle under the action of possible constant electric and magnetic fields. Which of the following are possible?
Answer
  1. $\text{E}=0,\ \text{B}\not=0$
Explanation:
A charged particle moves along a circle that mean Magnetic force is provides centripetal force that causes particle is move in a circle.
So, $\text{E}=0,\ \text{B}\not=0$
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Question 271 Mark
A cubical region of space is filled with some uniform electric and magnetic fields. An electron enters the cube across one of its faces with velocity v and a positron enters via opposite face with velocity -v. At this instant,
Answer
  1. The magnetic forces on both the particles cause equal accelerations.
  2. Both particles gain or loose energy at the same rate.
  3. The motion of the centre of mass (CM) is determined by B alone.
Here, magnitude of acceleration of electron and positron is same but direction is different.
Hence, option (a) is not correct.
Here,
Magnetic force on electron, $\vec{\text{F}}=-\text{e}(\vec{\text{v}}\times\vec{\text{B}})$
Magnetic force on positron, $\vec{\text{F}'}=-\text{e}(\vec{\text{v}}\times\vec{\text{B}})$
Acceleration of electron, $\vec{\text{a}}=\frac{\vec{\text{F}}}{\text{m}}=-\text{e}\frac{\big(\vec{\text{v}}\times\vec{\text{B}}\big)}{\text{m}}$
Acceleration of positron, $\vec{\text{a}'}=\frac{\big(\vec{\text{v}}\times\vec{\text{B}}\big)}{\text{m}}$
Now,
Net magnetic force on electron positron pair $=-2\text{e}\big(\vec{\text{v}}\times\vec{\text{B}}\big)$
Net electric on electron-positron pair = 0.
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Question 281 Mark
A proton beam is going from north to south and an electron beam is going from south to north. Neglecting the earth's magnetic field, the electron beam will be deflected
Answer
  1. Towards the proton beam.
Explanation:
A proton beam is going from north to south, so the direction of the current due to the beam will also be from north to south. Also, an electron beam is going from south to north, so the direction of the current due to the beam will also be from north to south. The direction of conventional current is along the direction of the flow of the positive charge and opposite to the flow of the negative charge. The magnetic field generated due to them will enter the plane of paper in the west and come out of the plane of paper in the east, according to the right-hand thumb rule. Since both the beams have currents in the same direction, they will apply equal and opposite forces on each other and, hence, will attract each other.
Thus, the electron beam will be deflected towards the proton beam.
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Question 291 Mark
A metallic rod of mass per unit length $0.5\ kg m^{–1}$ is lying horizontally on a smooth inclined plane which makes an angle of $30^\circ$ with the horizontal. The rod is not allowed to slide down by flowing a current through it when a magnetic field of induction $0.25$ T is acting on it in the vertical direction. The current flowing in the rod to keep it stationary is:
Answer
$11.32\  A.$
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Question 301 Mark
Two charged particles traverse identical helical paths in a completely opposite sense in a uniform magnetic field.
Answer
  1. The charge to mass ratio satisfy: $\Big(\frac{\text{e}}{\text{m}}\Big)_1+\Big(\frac{\text{e}}{\text{m}}\Big)_2=0$.
Solution:
Key concept: In this situation if the particle is thrown in x-y plane (as shown in figure) at some angle θ with velocity v, then we have to resolve the velocity of the particle in rectangular components, such that one component is along the field (v cosθ) and other one is perpendicular to the field (v sinθ). We find that the particle moves with constant velocity v cosθ along the field. The distance covered by the particle along the magnetic field is called pitch.

The pitch of the helix, (i.e., linear distance travelled in one rotation) will be given by
$\text{p}=\text{T}(\text{v}\cos\theta)=2\text{p}\frac{\text{m}}{\text{qB}}(\text{v}\cos\theta)$
For given pitch p correspond  to charge particle, we have
$\frac{\text{q}}{\text{m}}=\frac{2\pi\text{v}\cos\theta}{\text{qB}}=\text{constant}$
Here in this case, charged particles traverse identical helical paths in a completely opposite sense in a uniform magnetic field B, LHS for two particles should be same and of opposite sign. Therefore,
$\Big(\frac{\text{e}}{\text{m}}\Big)_1+\Big(\frac{\text{e}}{\text{m}}\Big)_2=0$
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Question 311 Mark
A positive charge is moving vertically upwards in the magnetic field towards the south. In which direction will it be deflected?
Answer
  1. East
Explanation:
If the positive charge is moving upwards the corresponding current is also upwards, the magnetic field is towards the south.
Using Flemings left-hand rule force is towards east and deflection towards east.
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Question 321 Mark
Cyclotron is a device used to _________.
Answer
  1. Accelerate the positively charged particles
Explanation:
A cyclotron is a device used to accelerate positively charged particles.
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Question 331 Mark
Identify the quantity which changes when a charged particle moves through a magnetic field?
Answer
  1. Direction of motion
Explanation:
When a charged particle moves through a magnetic field it suffers a change in its direction of motion.
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Question 341 Mark
A long, straight wire of radius $R$ carries a current distributed uniformly over its cross section. $T$ he magnitude of the magnetic field is:
Answer
Minimum at the axis of the wire.
Maximum at the surface of the wire.
A long, straight wire of radius $R$ is carrying current $i,$ which is uniformly distributed over its cross section.
According to Ampere's law,
$\oint\overrightarrow{\text{B}}.\text{d}\overrightarrow{\text{l}}=\mu_0\text{i}_\text{inside}$
At surface,
$\text{B}\times2\pi\text{R}=\mu_0\text{i}$
$\Rightarrow\text{B}_\text{surface}=\frac{\mu_0\text{i}}{2\pi\text{R}}$
Inside, $\text{B}\times2\pi\text{r}=\mu_0\text{i}$ for $ \text{r}<\text{R}$
Here $i^,$ is the current enclosed by the amperian loop drawn inside the wire.
$B_\text{inside}$ will be proportional to the distance from the axis.
On the axis
$B = 0$
The magnetic fields from points on the cross section will point in opposite directions and will cancel each other at the centre.
Outside, $\text{B}\times2\pi\text{r}=\mu_0\text{i}$
$\Rightarrow\text{B}_\text{outside}=\frac{\mu_0\text{i}}{2\pi\text{r}},\ \text{r}>\text{R}$
Therefore, the magnitude of the magnetic field is maximum at the surface of the wire and minimum at the axis of the wire.
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Question 351 Mark
Ampere’s circuital law states that:
Answer
  1. The line integral of magnetic field along the boundary of the open surface is equal to μ0 times the total current passing through the surface.
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Question 361 Mark
A long solenoid has a radius a and number of turns per unit length n. If it carries a current i, then the magnetic field on its axis is directly proportional to:
Answer
  1. $\text{ni}$
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Question 391 Mark
A current carrying loop is placed in a uniform magnetic field. The torque acting on it does not depend upon:
Answer
  1. Shape of the loop.
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Question 401 Mark
A current carrying power line carries current from west to east. Then the direction of the magnetic field 2m above it is:
Answer
  1. North to south
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Question 411 Mark
Consider three quantities $\text{x}=\frac{\text{E}}{\text{B}},\ \text{y}=\sqrt{\frac{1}{\mu_0\epsilon_0}}$ and $\text{z}=\frac{1}{\text{CR}}.$ Here, l is the length of a wire, C is a capacitance and R is a resistance. All other symbols have standard meanings.
Answer
  1. x, y have the same dimensions.
  2. y, z have the same dimensions.
  3. z, x have the same dimensions.
Explanation:
Lorentz Force:
$\text{qvB}=\text{qE}$
⇒ Dimensions of $\text{x}=[\text{v}]=\Big[\frac{\text{E}}{\text{B}}\Big]=\big[\text{LT}^{-1}\big]$
$\text{y}=\frac{1}{\sqrt{\mu_0\epsilon_0}}=\sqrt{\frac{4\pi}{\mu_0}\times\frac{1}{4\pi\epsilon_0}}=\sqrt{\frac{9\times10^9}{10^{-7}}}=3\times10^8=\text{c}$
⇒ Dimensions of $\text{y}=[\text{c}]=\big[\text{LT}^{-1}\big]$
Time constant of RC circuit = RC so dimensionally [RC] = [T]
$\Rightarrow\text{z}=\Big[\frac{\text{l}}{\text{RC}}\Rightarrow[\text{z}]=[\text{LT}^{-1}]$
Therefore, x, y and z have the same dimensions.
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Question 421 Mark
The strength of the magnetic field around an infinite current carrying conductor is:
Answer
  1. Inversely proportional to the distance
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Question 431 Mark
What is the work done by the magnetic field on the moving charge?
Answer
  1. No work is done by the magnetic field on the moving charge.
Explanation:
The magnetic force acts in such a way that the direction of the magnetic force and velocity are always perpendicular to each other. If force and velocity are perpendicular force and displacement are also perpendicular, thus W = FS cos q, if q = 90, work done will be zero.
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Question 441 Mark
A vertical straight conductor carries a current vertically upwards. A point P lies to the east of it at a small distance and another point Q lies to the west at the same distance the magnetic field at P is:
Answer
  1. Same as at Q.
Explanation:
As per Biot Savart Law, Magnetic field at a point is inversely proportional to square of distance from the current carrying conductor.
Therefore magnitude of magnetic field is same at both points P and Q, irrespective of their position from the conductor.
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Question 451 Mark
A charged particle goes undeflected in a region containing an electric and a magnetic field. It is possible that
Answer
  1. $\overrightarrow{\text{E}}||\overrightarrow{\text{B}},\ \overrightarrow{\text{v}}||\overrightarrow{\text{E}}$
  2. $\overrightarrow{\text{E}}$ is not parallel to $\overrightarrow{\text{B}}$
Explanation:
$\Rightarrow\overrightarrow{\text{V}}\overrightarrow{\text{E}},\ \overrightarrow{\text{B}}\overrightarrow{\text{E}}$
In this case Magnetic force on the particle is zero & $\overrightarrow{\text{V}}$ is paralle to $\overrightarrow{\text{E}}.$ So charged particle goes undeflected in a region.
$\overrightarrow{\text{E}}$ is not parallel to $\overrightarrow{\text{B}},$ But $\overrightarrow{\text{V}}$ is parallel to $\overrightarrow{\text{E}}.$
 
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Question 461 Mark
 What is the magnetic field inside a pipe?
Answer
  1. Zero
Explanation:
The magnetic field inside a pipe, i.e. inside a hollow cylindrical wire is zero.
This is due to the symmetry of the situation (pipe).
The pipe can be considered as a series of thin wires arranged in a circle.
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Question 471 Mark
When the free ends of a tester are dipped into a solution, the magnetic needle shows deflection.
Answer
  1. True
Explanation:
The deflection in the compass needle shows that current is flowing through the wounded wire and hence, through the circuit.
The circuit is complete since free ends of the tester are dipped in a solution.
The solution is certainly a conducting solution. That is the reason why the compass needle shows a deflection.
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Question 481 Mark
A charged particle moves through a magnetic field in a direction perpendicular to it. Then the:
Answer
  1. Speed of the particle remains unchanged.
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Question 491 Mark
In a coaxial, straight cable, the central conductor and the outer conductor carry equal currents in opposite directions. The magnetic field is zero:
Answer
  1. Outside the cable.
  2. Inside the inner conductor.
Explanation:
According to Ampere's law, in a coaxial, straight cable carrying currents i in the inner conductor and -i (equally in the opposite direction) in the outside conductor.
Inside the inner conductor
$\oint\overrightarrow{\text{B}}.\text{d}\overrightarrow{\text{l}}=\mu_0\text{i}_\text{inside}$
$\oint\overrightarrow{\text{B}}.\text{d}\overrightarrow{\text{l}}=0$
$\Rightarrow\text{b.l}=0$
$\Rightarrow\text{B}=0$
In between the 2 conductors
$\oint\overrightarrow{\text{B}}.\text{d}\overrightarrow{\text{l}}=\mu_0\text{i}$
$\Rightarrow\text{B}=\frac{\mu_0\text{i}}{2\pi\text{r}}$
Outside the outer conductor
$\oint\overrightarrow{\text{B}}.\text{d}\overrightarrow{\text{l}}=\mu_0\text{i}$
$\Rightarrow\text{B}=0$
Therefore, the magnetic field is zero outside the cable.
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Question 501 Mark
Two particles $X$ and $Y$ having equal charge, after being accelerated through the same potential difference enter a region of 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:
Answer
Particles $X$ and $Y$ of respective masses $m_1$ and $m_2$ are carrying charge q describing circular paths with respective radii $R_1$ and $R_2$ such that,$\text{R}_1=\frac{\text{m}_1\text{v}_1}{\text{qB}}$
$\text{R}_1=\frac{\text{m}_2\text{v}_2}{\text{qB}}$
Since both the particles are accelerated through the same potential difference, both will have the same kinetic energy.
$\therefore\frac{1}{2}\text{m}_1\text{v}_1^2=\frac{1}{2}\text{m}_2\text{v}_2^2$
$\because\text{R}_1=\frac{\text{m}_1\text{v}_1}{\text{qB}}$
$\Rightarrow\text{v}_1=\frac{\text{R}_1\text{qB}}{\text{m}_1}$
And,
$\text{R}_2=\frac{\text{m}_2\text{v}_2}{\text{qB}}$
$\Rightarrow\text{v}_2=\frac{\text{R}_2\text{qB}}{\text{m}_2}$
$\therefore\text{m}_1\Big(\frac{\text{R}_1\text{qB}}{\text{m}_1}\Big)^2=\text{m}_2\Big(\frac{\text{R}_2\text{qB}}{\text{m}_2}\Big)^2$
$\Rightarrow\frac{\text{m}_1}{\text{m}_2}=\frac{\text{R}_1^2}{\text{R}_2^2}=\Big(\frac{\text{R}_1}{\text{R}_2}\Big)^2$
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M.C.Q (1 Marks) - Physics STD 12 Science Questions - Vidyadip