Moving Charges and Magnetism Test

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Moving Charges and Magnetism Test - 67

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Question 1
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A moving coil galvanometer of resistance $$100 \space \Omega$$ is used as an ammeter using a resistance $$0.1 \space \Omega$$. The maximum deflection current in the galvanometer is $$100 \space \mu A$$.Find the minimum current in the circuit so that the ammeter shows maximum deflection.

A
$$100.1 mA$$

B
$$1000.1 mA$$

C
$$10.01 mA$$

D
$$1.01 mA$$

Solution

$$I_g G = \left (I - I_g \right )S$$.
Here, $$I_g = 100 \times 10^{-6} A$$
$$G = 100 \space \Omega, S = 0.1 \space \Omega$$

$$\therefore I = I_g \left ( \dfrac{G} {S} + 1 \right ) = 100 \times 10^{-6} \left ( \dfrac{100} {0.1} + 1 \right )$$
$$ = 100 \times 10^{-6} \times 1000.1 = 100.01 mA$$
Question 2
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A magnetic field $$ \vec{B} = B_{0}\hat{j} $$ exists in the region $$ a < x < 2a $$ and $$ \vec{B} = -B_{0}\hat{j} $$ , in the region $$ 2a < x < 3a $$ , where $$ B)_{0} $$ is a positive constant . A positive point charge moving with a velocity $$ \vec{v} = v_{0} \hat{i} $$ , where $$ v_{0} $$ is a positive constant , enters the magnetic field at $$ x = a $$ . The trajectory of the charge in this region can be like

A

B

C

D

Solution

Force experienced by charge q is F = q(V × B)
In the region, x = a to x = 2a,
$$F_1 = q(V_0i × B_0j) = qV_0B_0 \ k$$…..(i × j = k)

F1 is along positive z axis. In the region, x = 2a to 3a,

$$ F_2 = q(V_0i × – B_0j) = – qV_0B_0\ k$$
F2 is directed along negative z axis

∵ F1 & F2 are perpendicular to velocity V.
Question 3
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Two infinite long wires, each carrying current I , are lying along x and y -axis , respectively .A charged particle, having a charge q and mass m, is projected with a velocity u along the straight line OP. the path of the particle is ( neglect gravity)

A
straight line

B
circle

C
helix

D
cycloid

Solution

$$\overrightarrow B$$ (magnetic field) at all the points lying on OP is zero because both the wire cancels the effect of each other so u will not be affected by the magnetic force. Hence, it moves on a straight line.
Question 4
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Bolt-savart law indicates that the moving electrons (velocity v) produce a magnetic field b such that

A
.$$B \bot v$$

B
B $$B || v.$$

C
it obeys inverse cube law.

D
it is along the line joining the electron and point of observation.

Solution

By Biot-savart law, $$ dB=\dfrac{I.dlsin\theta }{r^{2}} $$
Or $$ dB= \dfrac{I\times dl}{r} $$
acc to Biot-Savart law , if magnetic field is not perpendicular to the motion of charge then it will not move in helical path, which is not possible for motion of a charge in magnetic field.
So the magnetic field is perpendicular to the direction of flow of charge verifies answer 'a'.
Question 5
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Two parallel wires are carrying electric currents of equal magnitude and in the same direction. They exert

A
An attractive force on each other

B
An repulsive force on each other

C
No force on each other

D
A rotational torque on each other

Solution

Two straight conductors carry current in same direction, then attractive force acts between them.
Question 6
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The radius of curvature of the path of the charges particles in a uniform magnetic field is directly proportional to

A
The charge on the particle

B
The momentum on the particle

C
The energy on the particle

D
The intensity of the field

Solution

The radius of the circular path in the magnetic field is given by:

$$r=\dfrac{mv}{qB}$$

We know that $$mv$$ is the momentum of the body.

$$\therefore r=\dfrac {p}{qB}$$

$$\Rightarrow r\propto p$$
Question 7
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Tick the most appropriate answer.
Which of the following methods is used to magnetize a magnetic without touching it by a magnet?

A
Single-touch

B
Double-touch

C
Using electric current

D
None of these

Solution

Using electric current we can magnetize a magnetic material without touching it by a magnet.
Question 8
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A magnetic field

A
Always exerts a force on a charged particle

B
Never exerts a force on a charged particle

C
Exerts a force, if the charges particle is moving across the magnitude field lines

D
Exerts a force, if the charges particle is moving along the magnitude field lines

Solution

The force exerted by the magnetic field is given by:
$$\vec F=q\vec v \times \vec B$$

So, the force is exerted only when the charge particle is moving across the magnetic field lines.
Question 9
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A proton (mass $$=1.67\times 10^{-27} kg$$ and charge $$=1.6\times 10^{-19}C$$) enters perpendicular to a magnetic field of intensity $$2$$ weber $$/m^2$$ with a velocity $$3.4\times 10^7\ m/sec$$. The acceleration of the proton should be

A
$$6.5\times 10^{15}m/sec^2$$

B
$$6.5\times 10^{13}m/sec^2$$

C
$$6.5\times 10^{11}m/sec^2$$

D
$$6.5\times 10^{9}m/sec^2$$

Solution

The force experienced by the proton due to the magnetic field provides an acceleration to it.
$$F=ma=qvB $$

$$\Rightarrow a=\dfrac {qvB}{m}$$

$$=\dfrac {1.6\times 10^{-19}\times 2\times 3.4\times 10^{7}}{1.67\times 10^{-27}}$$

$$=6.5\times 10^{15}\ m/sec$$
Question 10
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Lorentx force can be calculate dby using the formula

A
$$\vec F=q(\vec E+\vec v\times \vec B)$$

B
$$\vec F=q(\vec E-\vec v\times \vec B)$$

C
$$\vec F=q(\vec E+\vec v . \vec B)$$

D
$$\vec F=q(\vec E\times \vec v+\vec B)$$

Solution

Lorentz force is given by
$$\vec F=\vec {F_e}+\vec {F_m}$$

$$=q\vec E+q(\vec v \times \vec B)$$

$$=q [\vec E+(\vec v\times \vec B)]$$