Moving Charges and Magnetism Test

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Moving Charges and Magnetism Test - 37

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Weekly Quiz Competition

Question 1
1 / -0

Two particles having same charge and $$KE$$ enter at right angles into the same magnetic field and travel in circular paths of radii $$2 cm$$ and $$3 cm$$ respectively. The ratio of their masses is :

A
$$2 : 3$$

B
$$3 : 2$$

C
$$4 : 9$$

D
$$9 : 4$$

Solution

Radius of charged particle in magnetic field is,
$$r=\dfrac{mv}{qB}$$
$$\dfrac{mv^{2}}{2}=K_{e}$$
$$r=\dfrac{\sqrt{2K_{e}m}}{qB}$$
$$r_{1}=\dfrac{\sqrt{2K_{e}m_{1}}}{qB}$$
$$r_{2}=\dfrac{\sqrt{2K_{e}m_{2}}}{qB}$$
$$\dfrac{4}{9}=\dfrac{m_{1}}{m_{2}}$$
Question 2
1 / -0

Two long straight parallel wires separated by a distance, carrying equal currents exert a force F per unit length on each other. If the distance of separation is doubled, and the current in each is halved, the force per unit length, between them will be :

A
F

B
F/2

C
F/4

D
F/8

Solution

$$F=\dfrac { \mu_o{ i }_{ 1 }{ i }_{ 2 } }{ 2\pi d } $$
by halfing each current, we get and distance is doubled, then

$$F=\dfrac { \mu_o\dfrac { { i }_{ 1 } }{ 2 } \quad \dfrac { { i }_{ 2 } }{ 2 } }{ 2\pi \quad (2d) } =\dfrac { 1 }{ 8 } \quad \bigg (\dfrac { \mu_o{ i }_{ 1 }{ i }_{ 2 } }{ 2\pi d } \bigg )=\dfrac { F }{ 8 } $$
Question 3
1 / -0

A proton of energy $$2 MeV$$ is moving perpendicular to uniform magnetic field of $$2.5 T$$. The force on the proton is : ($$M_p=1.6\times 10^{-27}kg$$ and $$q_{p}=e=1.6\times 10^{-19}C$$ )

A
$$2.5\times 10^{-16}N$$

B
$$8\times 10^{-11}N$$

C
$$2.5\times 10^{-11}N$$

D
$$8\times 10^{-12}N$$

Solution

$$F=qvB$$
$$\dfrac{mv^{2}}{2}=K_{e}$$
$$v=\sqrt{\dfrac{2K_{e}}{m}}$$
$$\therefore f=q\times \sqrt{\dfrac{2K_{e}}{m}}\times B$$
=$$1.6\times 10^{-19}\times \sqrt{\dfrac{2\times 2\times 10^{6}\times 1.6\times 10^{-19}}{1.6\times 10^{-27}}}\times 2.5$$
=$$8\times 10^{-12} N$$
Question 4
1 / -0

A proton of energy $$E$$ is moving along a circular path in a uniform magnetic field. If an alpha particle describes the same circular path, its energy should be :

A
$$4E$$

B
$$2E$$

C
$$E$$

D
$$0.5E$$

Solution

proton $$\dfrac{mv^{2}}{2}=E$$
$$r=\dfrac{mv}{qB}$$=$$\dfrac{\sqrt{2Em}}{qB}$$
for $$\alpha $$ particle, $$M=4m \ and \ Q=2q$$

$$r=\dfrac{\sqrt{2{E}'4m}}{2qB}$$

$$r=\dfrac{\sqrt{2{E}'m}}{qB}$$

$${E}'=E$$
Question 5
1 / -0

An electron travelling with a velocity $$\bar{V}=10^{7}i\ m/s$$ enter a magnetic field of induction $$\bar{B}=\overline{2j}$$ . The force on electron is

A
$$1.6\times 10^{-12}\bar{j}N$$

B
$$3.2\times 10^{-12}\bar{k}N$$

C
$$6.4\times 10^{-12}\bar{k}N$$

D
$$-3.2\times 10^{-12}\bar{k}N$$

Solution

$$\bar{F}=q\left ( \underset{V}{\rightarrow}\times \underset{B}{\rightarrow} \right )$$
=$$q\times 10^{7}\times 2\times \bar{k}$$
=$$1.6\times 10^{-19}\times 2\times 10^{7} \bar{k} N$$
=$$3.2\times 10^{-12}\bar{k} N$$
Question 6
1 / -0

Electrons accelerated by a potential difference $$V$$ enter a uniform magnetic field of flux density $$B$$ at right angles to the field. They describe a circular path of radius $$r$$. If now $$V$$ is doubled and $$B$$ is also doubled, the radius of the new circular path is :

A
$$4r$$

B
$$2r$$

C
$$2\sqrt{2r}$$

D
$$\dfrac{r}{\sqrt{2}}$$

Solution

Radius of charged particle in magnetic field is,
$$r=\dfrac{mv}{qB}$$
$$ \dfrac{mv^{2}}{2}=qV$$
$$mv=\sqrt{2qVm}$$
$$r=\dfrac{mv}{qB}=\dfrac{\sqrt{2qVm}}{qB}=\dfrac{1}{B}\times\bigg (\sqrt{\dfrac{2Vm}{q}}\bigg )$$
$${r}'=\dfrac{1}{2B}\times\bigg (\sqrt{\dfrac{2\times 2Vm}{q}}\bigg )=\dfrac{r}{\sqrt{2}}$$
Question 7
1 / -0

A proton moving with a velocity of $$(6\hat{i} + 8\hat{j}) \times 10^{5} \ ms^{-1}$$ enters uniform magnetic field of induction $$5 \times 10^{-3} \hat{k}T$$ . The magnitude of the force acting on the proton is :
($$\hat{i}, \hat{j}$$ and $$\hat{k}$$ are unit vectors forming a right handed triad)

A
$$0$$

B
$$8\times 10^{-16}N$$

C
$$3\times 10^{-16}N$$

D
$$4\times 10^{-16}N$$

Solution

Force on a charged particle in magnetic field is given by,
$$\bar{F}=q\left ( {\vec{V}}{}\times \vec{B}{} \right )$$
=$$1.6\times 10^{-19}((6\hat{i}+8\hat{j})\times 5\hat{k})\times10^{5}\times 10^{-3}$$
=$$1.6\times 10^{-19}\times 10^{-2}(-30\hat{j}+40\hat{i})$$
=$$1.6\times 10^{-16}(-3\hat{j}+4\hat{i})$$
$${\left | F \right |}=1.6\times 10^{-16}\sqrt{3^{2}+4^{2}}$$
=$$1.6\times 10^{-16}\times 5 N$$
=$$8\times 10^{-16} N$$
Question 8
1 / -0

If a particle of charge $$10^{-12}C$$ moving along the x-axis with a velocity $$10^{5}$$ m/s. experiences a force of $$10^{-10}N$$ in y-direction due to magnetic field, then the minimum magnetic field is.

A
$$6.25\times 10^{3}T$$ in Z direction

B
$$10^{-15}T$$ in Z direction

C
$$6.25\times 10^{-3}T$$ in Z direction

D
$$10^{-3}T$$ in -ve Z direction

Solution

$$f=qVB\sin \theta $$

B will be minimum when $$\theta =90^{\circ}$$
$$B=\dfrac{f}{qV}$$
=$$\dfrac{10^{-10}}{10^{-12}\times 10^{5}}$$
=$$10^{-3}T$$ in Z-direction
Question 9
1 / -0

Acceleration experienced by a particle with specific charge $$1 \times10^{7} C/kg$$ when fired perpendicular to a magnetic field of induction $$100 T$$ with a velocity $$10^{5} ms^{-1}$$ is :

A
$$10^{8} ms^{-2}$$

B
$$10^{-6} ms^{-2}$$

C
$$10^{14} ms^{-2}$$

D
$$10^{-8} ms^{-2}$$

Solution

Specific charge $$\left(\dfrac{q}{m}\right)=10^{7} C/kg$$
$$\left | B \right |=100T$$

$$f=qVB \sin \theta$$

$$q=\dfrac{f}{m} $$

=$$\dfrac{qVB \sin \theta}{m} $$

=$$\dfrac{10^{7}\times 10^{5}\times 100 \times \sin 90^{\circ}}{1}$$

=$$10^{14} ms^{-2}$$
Question 10
1 / -0

An $$\alpha $$ -particle describes a circular path of radius $$r$$ in a magnetic field $$B$$. The radius of the circular path described by the proton of same energy in the same magnetic field is :

A
$$\dfrac{r}{2}$$

B
$$r$$

C
$$\sqrt{2}r$$

D
$$2r$$

Solution

For $$ \ \alpha-$$ particle,
$$M=4m$$
$$q=2e$$
So,$$r=\dfrac{\sqrt{2K_{e}M}}{qB}$$
=$$\dfrac{\sqrt{2K_{e}m}}{eB}$$

For proton $$M=m$$ ,
$$q=e$$
$$\therefore R=\dfrac{Mv}{qB}$$
=$$\dfrac{\sqrt{2K_{e}M}}{qB}$$
=$$\dfrac{\sqrt{2K_{e}m}}{eB}$$
$$=r$$

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Re-Attempt Weekly Quiz Competition

Moving Charges and Magnetism Test - 37

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Moving Charges and Magnetism Test - 37

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Question 1

$$2 : 3$$

$$3 : 2$$

$$4 : 9$$

$$9 : 4$$

Solution

Question 2

F

F/2

F/4

F/8

Solution

Question 3

$$2.5\times 10^{-16}N$$

$$8\times 10^{-11}N$$

$$2.5\times 10^{-11}N$$

$$8\times 10^{-12}N$$

Solution

Question 4

$$4E$$

$$2E$$

$$E$$

$$0.5E$$

Solution

Question 5

$$1.6\times 10^{-12}\bar{j}N$$

$$3.2\times 10^{-12}\bar{k}N$$

$$6.4\times 10^{-12}\bar{k}N$$

$$-3.2\times 10^{-12}\bar{k}N$$

Solution

Question 6

$$4r$$

$$2r$$

$$2\sqrt{2r}$$

$$\dfrac{r}{\sqrt{2}}$$

Solution

Question 7

$$0$$

$$8\times 10^{-16}N$$

$$3\times 10^{-16}N$$

$$4\times 10^{-16}N$$

Solution

Question 8

$$6.25\times 10^{3}T$$ in Z direction

$$10^{-15}T$$ in Z direction

$$6.25\times 10^{-3}T$$ in Z direction

$$10^{-3}T$$ in -ve Z direction

Solution

Question 9

$$10^{8} ms^{-2}$$

$$10^{-6} ms^{-2}$$

$$10^{14} ms^{-2}$$

$$10^{-8} ms^{-2}$$

Solution

Question 10

$$\dfrac{r}{2}$$

$$r$$

$$\sqrt{2}r$$

$$2r$$

Solution

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