Moving Charges and Magnetism Test

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Moving Charges and Magnetism Test - 21

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Question 1
1 / -0

Two long straight parallel wires, carrying (adjustable) current $$I_1$$ and $$I_2$$, are kept at a distance d apart. If the force 'F' between the two wires is taken as 'positive' when the wires repel each other and 'negative' when the wires attract each other, the graph showing the dependence of 'F', on the product $$I_1I_2$$, would be

A

B

C

D

Solution

The force between two wires per unit length is: $$F=\dfrac{\mu _0}{2 \pi d} I_1I_2$$

When,$$I_1$$ and $$I_2$$ are in in the same direction, the wires attract each other, and $$F \lt 0$$
When,$$I_1$$ and $$I_2$$ are in opposite direction, the wires repel each other, and $$F \gt 0$$
The plot is a straight line since $$F \propto I_1I_2$$
The slop is negative since sign of force is opposite to the product $$I_1I_2$$( same direction product is positive, opposite direction product is negative.
Question 2
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An electron gun is placed inside a long solenoid of radius $$R$$ on its axis. The solenoid has $$n$$ tums / length and carries a current $$I$$. The electron gun shoots an electron along the radius of the solenoid with speed $$\upsilon $$. If the electron does not hit the surface of the solenoid, maximum possible value of $$\upsilon $$ is (allsymbols have their standard meaning):

A
$$\dfrac {e \mu_0 nIR}{4m}$$

B
$$\dfrac {e \mu_0 nIR}{m}$$

C
$$\dfrac {2e \mu_0 nIR}{m}$$

D
$$\dfrac {e \mu_0 nIR}{2m}$$

Solution

Magnetic field inside a solenoid, $$B = (\mu_0 nI)$$
Since, velocity V is given to the electron.
$$q = e = $$ charge on an electron.
$$m = $$ mass of an electron.
Magnetic force, $$F = q \vee B$$ Ref. image I
$$F = (eV \mu_0 nI)$$

In a magnetic field, charge will move on a circular path. Here, If radius is just less than or equal to radius of solenoid then the electron does not hit the surface.
$$\therefore F = q B \vee = \dfrac{mV^2}{R}$$ Ref. image II
$$\Rightarrow mv = q BR$$ $$\dfrac{mV^2}{R}$$ is centripetal force
$$\Rightarrow v = \dfrac{(qR)B}{m}$$
$$q = e$$
$$\Rightarrow v = \dfrac{qR \times \mu_0 nI}{m} = \dfrac{e \mu_0 nIR}{m}$$
$$\therefore V = \dfrac{e \mu_0 n IR}{m}$$ maximum possible value of V.
Option (2) is correct.
Question 3
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An electron, a proton and an alpha particle having the same kinetic energy are moving in circular orbits of radii $$r_e, r_p, r_{\alpha}$$ respectively in a uniform magnetic field B. The relation between $$r_e, r_p, r_{\alpha}$$ is?

A
$$r_e < r_p < r_{\alpha}$$

B
$$r_e < r_{\alpha} < r_p$$

C
$$r_e > r_p = r_{\alpha}$$

D
$$r_e < r_p = r_{\alpha}$$

Solution

We know that $$r = \dfrac{mv}{qB}$$
Also $$ mv = \sqrt{2Km}$$
where K is the kinetic energy
Hence $$r = \dfrac{\sqrt{2Km}}{qB}$$
Since $$K$$ and $$B$$ are constants,
$$r = K' \sqrt{m}{B}$$
Since $$m_{\alpha}= 4 m_{p}$$
and $$q_{\alpha}= 2 q_{p}$$,
$$r_{\alpha}= r_{p}$$
and Since $$m_{p}> m_{e}$$
and $$q_{p}= q_{e}$$
$$r_{p}>r_{e}$$
Question 4
1 / -0

A negative test charge is moving near a long straight wire carrying a current. The force acting on the test charge is parallel to the direction of the current. The motion of the charge is :

A
Away from the wire

B
Parallel to the wire along the current

C
Towards the wire

D
Parallel to the wire opposite to the current

Solution

Magnetic field at A $$\rightarrow$$ out of the plane of the paper
$$B={ B }_{ 0 }\hat { k }$$
$$\Rightarrow$$ force on charged particle:
$$\Rightarrow F=-q\vec { v } \times \vec { B } $$
$$\Rightarrow F=+qv{ B }_{ 0 }\hat { j } $$
Therefore, the particle moves towards the wire.
Question 5
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An electron traveling with a speed $$\mathrm{u}$$ along the positive $$\mathrm{x}$$-axis enters into a region of magnetic field where $$\mathrm{B}=-\mathrm{B}_{0}\hat{\mathrm{k}}(\mathrm{x}>0)$$. It comes out of the region with speed $$\mathrm{v}$$ then :

A
v = u at y > 0

B
v = u at y < 0

C
v > u at y > 0

D
v > u at y < 0

Solution

Charged particle moves in a circular trajectory with constant speed $$u$$, since force always acts perpendicular to the motion. It is not able to change its speed. It just tends to come out after one circle is completed with speed $$u$$.
Question 6
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$$\mathrm{A}$$ magnetic field $$\vec{\mathrm{B}}=\mathrm{B}_{0}\hat{\mathrm{j}}$$ exists in the region a $$<\mathrm{x}<2\mathrm{a}$$ and $$\vec{\mathrm{B}}=-\mathrm{B}_{0}\hat{\mathrm{j}}$$ , in the region $$2\mathrm{a}<\mathrm{x}<3\mathrm{a}$$, where $$\mathrm{B}_{0}$$ is a positive constant. $$\mathrm{A}$$ positive point charge moving with a velocity $$\vec{\mathrm{v}}=\mathrm{v}_{0}\hat{\mathrm{i}}$$, where $$\mathrm{v}_{0}$$ is a positive constant, enters the magnetic field at $$\mathrm{x}=\mathrm{a}$$. The trajectory of the charge in this region can be like,

A

B

C

D

Solution

$$F=q vB$$
Now $$\bar v = v_0 \hat i , \bar B = B_0 \hat j Thus, \bar F = v_0B_0 \hat k$$
Thus the particle will start straying towards +z direction.
After it reaches x=2a, $$\bar F = - v_0B_0 \hat k$$
There won't be any jerks as force always acts perpendicular to particle motion, and acceleration is finite.
Now it starts to stray towards -z direction.
Question 7
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A proton and alpha particle both enter a region of uniform magnetic field B, moving at right angles to the field B. If the radius of circular orbits for both the particles is equal and the kinetic energy acquired by proton is 1 MeV, the energy acquired by the alpha particle will be:

A
1 Me V

B
4 Me V

C
0.5 MeV

D
1.5 MeV

Solution

The radius of circular path is $$r=\dfrac{\sqrt{2mE}}{qB}$$
So, $$\dfrac{r_{\alpha}}{r_p}=\sqrt{\dfrac{m_{\alpha}E_{\alpha}}{m_pE_p}}\dfrac{q_p}{q_{\alpha}}$$

or, $$1=\sqrt{\dfrac{E_{\alpha}}{E_p}}$$ as radius of both particles are equal and the mass of alpha particle is 4 times of proton and charge is 2 times of proton.
Thus, $$E_{\alpha}=E_p=1 $$MeV
Question 8
1 / -0

Two identical bar magnets are fixed with their centres at a distance d apart. A stationary charge Q is placed at P in between the gap of the two magnets at a distance D from the centre O as shown in the figure.
The force on the charge Q is

A
$$zero$$

B
directed along OP

C
directed along PO

D
directed perpendicular to the plane of paper

Solution

Magnetic field due to bar magnets exerts force on moving charges only.
Since the charge is at rest, zero force acts on it.
Question 9
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Current sensitivity of a moving coil galvanometer is $$5$$ div/mA and its voltage sensitivity (angular deflection per unit voltage applied) is $$20$$ div/V. The resistance of the galvanometer is?

A
$$250$$ $$\Omega$$

B
$$40$$ $$\Omega$$

C
$$500$$ $$\Omega$$

D
$$25$$ $$\Omega$$

Solution

Current sensitivity

$$I_g=\dfrac{NBA}{K}=\dfrac{5div}{mA}=5000\displaystyle\frac{div}{A}$$

Voltage sensitivity $$v_s=\dfrac{NBA}{KR}=20\dfrac{div}{v}$$

$$\Rightarrow\displaystyle\frac{I_g}{v_s}=R$$

$$\Rightarrow R=\displaystyle\frac{5000}{20}=250\Omega$$

Option 1 is correct.
Question 10
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A proton carrying $$1MeV$$ kinetic energy is moving in a circular path of radius $$R$$ in uniform magnetic field. What should be the energy of an $$\alpha$$-particle to describe a circle of same radius in the same field?

A
$$1MeV$$

B
$$0.5MeV$$

C
$$4MeV$$

D
$$2MeV$$

Solution

For a particle of mass m and charge Q in a magnetic field, the radius of the path is given by,

$$r=\cfrac{\sqrt 2m(KE)}{qB}$$

$$q\propto \sqrt {m(KE)}$$

$$\cfrac{e}{2e}=\sqrt {\cfrac{({m}_{p})(IMeV)}{(4{m}_{p})(KE)}}$$

$$\cfrac{1}{4}=\cfrac{1}{4(KE)}$$

$$KE=1MeV$$