Moving Charges and Magnetism Test

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Moving Charges and Magnetism Test - 56

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Question 1
1 / -0

A proton of charge $$e$$ moving at speed $$v_0$$ is placed midway between two parallel wires a distance $$a$$ apart, each carrying current II in opposite directions.
The force on the proton is:

A
$$0$$

B
$$ev_0\cfrac{\mu_0I}{2\pi a}$$

C
$$ev_0\cfrac{\mu_0I}{2\pi (2a)}$$

D
$$2ev_0\cfrac{\mu_0I}{2\pi \frac{a}{2}}$$

E
Unable to be determined
Question 2
1 / -0

A cart with frictionless wheels and an attached loop of wire moves at constant velocity into a magnetic field that is perpendicular to the cart as shown.
Which of the following will occur?

A
Current is induced clockwise.

B
No current is induced because the velocity is constant.

C
Current is induced counterclockwise.

D
No current is induced because the loop is perpendicular to the magnetic field.

Solution

When loop enters in the magnetic field , the magnetic flux linked with loop changes therefore a current will induce in loop wire , doesn't matter velocity is constant or not .
Now according to Lenz's law the direction of current in loop will be such that it opposes the motion of loop in magnetic field i.e. force due to magnetic field on loop arm should be toward left , therefore by Fleming's left hand rule the current in loop will be clockwise .
Question 3
1 / -0

Pick out the WRONG statement.

A
The gain in the K.E. of the electron moving at right angles to the magnetic field is zero

B
When an electron is shot at right angles to the electric field, it traces a parabolic path

C
An electron moving in the direction of the electric field gains K.E.

D
An electron at rest experiences no force in the magnetic field

Solution

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Question 4
1 / -0

A proton beam enters a magnetic field of $$10^{-4}Wb\ m^{-2}$$ normally. If the specific charge of the proton is $$10^{11}C\ kg^{-1}$$ and its velocity is $$10^{9}\ ms^{-1}$$, then the radius of the circle described will be

A
$$0.1\ m$$

B
$$10\ m$$

C
$$100\ m$$

D
$$1\ m$$

Solution

Given : $$B = 10^{-4}$$ $$Wb/m^2$$ $$q = 10^{11}C/kg$$ $$v = 10^9 m/s$$

Radius of circle described, $$r = \dfrac{mv}{Bq}$$
$$\therefore$$ $$r = \dfrac{10^9}{10^{-4} \times 10^{11}} = 100\ m$$
Question 5
1 / -0

The Biot Savart's Law in vector form is

A
$$\overline {\delta B} = \dfrac {\mu_{0}}{4\pi} \dfrac {dl(\vec {l}\times \vec {r})}{r^{3}}$$

B
$$\overline {\delta B} = \dfrac {\mu_{0}}{4\pi} \dfrac {I(\vec {dl}\times \vec {r})}{r^{3}}$$

C
$$\overline {\delta B} = \dfrac {\mu_{0}}{4\pi} \dfrac {I(\vec {r}\times \vec {dl})}{r^{3}}$$

D
$$\overline {\delta B} = \dfrac {\mu_{0}}{4\pi} \dfrac {I(\vec {dl}\times \vec {r})}{r^{2}}$$

Solution

The Biot Savart's Law we know is given by:
$$\vec{\delta B} = \dfrac{\mu_0}{4\pi}\dfrac{I dl \sin(\theta)}{r^2}\hat{r}$$, where $$\theta$$ is the angle between the line element $$\vec{dl}$$ and the radial unit vector $$\hat{r}$$.
Now we know: $$\hat{r} = \dfrac{\vec{r}}{r}$$, using this and the cross product of two vectors we get; $$\vec{\delta B} = \dfrac{\mu_0}{4\pi}\dfrac{I(\vec{dl}\times \vec{r})}{r^3}$$
Question 6
1 / -0

An electron enters an electric field having intensity $$\vec { E } =3\hat { i } +6\hat { j } +2\hat { k } $$ $$V{m}^{-1}$$ and magnetic field having induction $$\vec { B } =2\hat { i } +3\hat { j } T$$ with a velocity $$\vec { V } =2\vec { i } +3\vec { j } $$ $${ms}^{-1}$$. The magnitude of the force acting on the electron is (Given $$e=-1.6\times {10}^{-19}C$$)

A
$$2.02\times {10}^{-18}N$$

B
$$5.16\times {10}^{-16}N$$

C
$$3.72\times {10}^{-17}N$$

D
$$4.41\times {10}^{-18}N$$

E
None of the above

Solution

The electric force acting on the electron is $$qE=-e(3\hat{i}+6\hat{j}+2\hat{k})N$$
The magnetic force acting on the electron=$$q(\vec{v}\times \vec{B})$$
$$=q((2\hat{i}+3\hat{j})\times(2\hat{i}+3\hat{j}))$$
$$=0$$
Hence net force acting on the electron is $$-e(3\hat{i}+6\hat{j}+2\hat{k})$$
Magnitude of this force=$$e(\sqrt{3^2+6^2+2^2})$$
$$=7e=11.2\times 10^{-19}N$$
Question 7
1 / -0

A metallic ring of radius $$a$$ and resistance $$R$$ is held fixed with its axis along a spatially uniform magnetic field whose magnitude is $$B_{0}\sin (\omega t)$$. Neglect gravity. Then;

A
The current in the ring oscillates with a frequency of $$2\omega$$

B
The joule heating loss in the ring is proportional to $$a^{2}$$

C
The force per unit length on the ring will be proportional to $$B_{0}^{2}$$

D
The net force on the ring is non-zero

Solution

According to Lenz's law, magnitude of induced current will be $$i=\dfrac{d\phi/dt}{R}$$
where $$\phi=B.A$$ where $$B = B_{0}\sin \omega t$$.
$$i=\dfrac{B_0 \omega cos \omega t \pi a^2}{R}$$
$$P=\int i^2Rdt \propto a^4$$
$$F=idl B$$
$$F\propto B_0^2$$
Question 8
1 / -0

A circular coil of $$200$$ turns and radius $$10 cm$$ is placed in an uniform magnetic field of $$0.1 T$$ normal to the plane of the coil. The coil carries a current of $$5 A$$. The coil is made up of copper wire of cross-sectional area $${10}^{-5}{m}^{2}$$ and the number of free electrons per unit volume of copper is $${10}^{29}$$. The average force experienced by an electron in the coil due to magnetic field is

A
$$5 \times {10}^{-25}N$$

B
Zero

C
$$8 \times {10}^{-24}N$$

D
None of these

Solution

$$V_d=\dfrac{i}{nqA}$$,

Force=$$F=q(V_d\times B)$$,

$$F= \dfrac{iB}{nA}=\dfrac{5\times0.1}{10^{29}\times10^{-5}}=5\times10^{-25}N$$
Question 9
1 / -0

A horizontal overhead power line carries a current of 90A in east to west direction. Magnitude of magnetic field due to the current 1.5 m below the line is :

A
$$1.2 T$$

B
$$1.2 \times 10^{-10} T$$

C
$$0T$$

D
$$1.2 \times 10^{-5} T$$

Solution

Magnitude of magnetic field at a distance $$r$$ due to straight current carrying conductor is given by
$$B=\dfrac{\mu_0}{2\pi}\dfrac{i}{r}$$
Given : $$i = 90$$ A and $$r = 1.5$$ m
$$B=2\times 10^{-7} \times \dfrac{ 90}{1.5}=1.2\times 10^{-5}T$$
Question 10
1 / -0

If flow of electric current is parallel to magnetic field, force will be:

A
Negative

B
Maximum

C
Zero

D
None of these

Solution

Electric current is basically flow of +ve charge (assumed) and force on a charge moving in any direction with velocity $$\vartheta$$ is given by
$$F=q(\vec{\vartheta }*\vec{B})$$
if electric current is parallel to $$\vec{B}$$ ,then $$\vec{\vartheta }*\vec{B}=0$$
$$\therefore F=0$$

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Moving Charges and Magnetism Test - 56

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Moving Charges and Magnetism Test - 56

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Question 1

$$0$$

$$ev_0\cfrac{\mu_0I}{2\pi a}$$

$$ev_0\cfrac{\mu_0I}{2\pi (2a)}$$

$$2ev_0\cfrac{\mu_0I}{2\pi \frac{a}{2}}$$

Unable to be determined

Question 2

Current is induced clockwise.

No current is induced because the velocity is constant.

Current is induced counterclockwise.

No current is induced because the loop is perpendicular to the magnetic field.

Solution

Question 3

The gain in the K.E. of the electron moving at right angles to the magnetic field is zero

When an electron is shot at right angles to the electric field, it traces a parabolic path

An electron moving in the direction of the electric field gains K.E.

An electron at rest experiences no force in the magnetic field

Solution

Question 4

$$0.1\ m$$

$$10\ m$$

$$100\ m$$

$$1\ m$$

Solution

Question 5

$$\overline {\delta B} = \dfrac {\mu_{0}}{4\pi} \dfrac {dl(\vec {l}\times \vec {r})}{r^{3}}$$

$$\overline {\delta B} = \dfrac {\mu_{0}}{4\pi} \dfrac {I(\vec {dl}\times \vec {r})}{r^{3}}$$

$$\overline {\delta B} = \dfrac {\mu_{0}}{4\pi} \dfrac {I(\vec {r}\times \vec {dl})}{r^{3}}$$

$$\overline {\delta B} = \dfrac {\mu_{0}}{4\pi} \dfrac {I(\vec {dl}\times \vec {r})}{r^{2}}$$

Solution

Question 6

$$2.02\times {10}^{-18}N$$

$$5.16\times {10}^{-16}N$$

$$3.72\times {10}^{-17}N$$

$$4.41\times {10}^{-18}N$$

None of the above

Solution

Question 7

The current in the ring oscillates with a frequency of $$2\omega$$

The joule heating loss in the ring is proportional to $$a^{2}$$

The force per unit length on the ring will be proportional to $$B_{0}^{2}$$

The net force on the ring is non-zero

Solution

Question 8

$$5 \times {10}^{-25}N$$

Zero

$$8 \times {10}^{-24}N$$

None of these

Solution

Question 9

$$1.2 T$$

$$1.2 \times 10^{-10} T$$

$$0T$$

$$1.2 \times 10^{-5} T$$

Solution

Question 10

Negative

Maximum

Zero

None of these

Solution

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