Moving Charges and Magnetism Test

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Moving Charges and Magnetism Test - 55

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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 $$I$$ in opposite directions.
The force on the proton is:

A
$$0$$

B
$$ev_0\cfrac{\mu_0 I}{2\pi a}$$

C
$$ev_0\cfrac{\mu_0 I}{2\pi (2a)}$$

D
$$ev_0\cfrac{\mu_0 I}{2\pi \dfrac{a}{2}}$$

E
Unable to be determined

Solution

In fact proton is under the effect of two equal magnetic fields due to two wires.
Let $$a$$ is the distance between two wires.
The magnetic field due to first wire carrying a current $$I$$ at a distance $$a/2$$ (midway) is given by:
$$B_{1}=\dfrac{\mu_{0}}{2\pi}.\dfrac{I}{a/2}$$
Magnetic field due to second wire carrying a current $$I$$ at a distance $$a/2$$ (midway) is given by:
$$B_{2}=\dfrac{\mu_{0}}{2\pi}.\dfrac{I}{a/2}$$
As the direction of current is opposite in two wires therefore by Right hand thumb rule, both the magnetic fields will be in the same direction, therefore force on proton due to two fields will be the sum of forces due to two fields.

$$F=ev_{0}B_{1}-ev_{0}B_{2}=ev_{0}\dfrac{\mu_{0}}{2\pi}.\dfrac{I}{a/2}-ev_{0}\dfrac{\mu_{0}}{2\pi}.\dfrac{I}{a/2}$$

or $$F=0$$
Question 2
1 / -0

A particle of charge $$q$$ and mass $$m$$ moves in a circular path of radius $$r$$ in a uniform magnetic field $$B$$.
The angular momentum of the particle can be described by the expression:

A
$$eBr$$

B
$$2eBr$$

C
$$eBr^2$$

D
$$eBrm_p$$

E
$$2eBrm_p$$

Solution

Angular momentum of the particle $$L = mv r$$
Using $$eB r = mv$$
$$\implies$$ $$L = (eBr ) r =eBr^2$$
Question 3
1 / -0

A particle of charge $$q$$ and mass $$m$$ is moving at a speed $$v$$ enters a uniform magnetic field of strength $$B$$ as shown below.
How much work is done by the magnetic field on the charge as the field accelerates the charge into a circle of radius $$r$$?

A
$$0$$

B
$$\cfrac{1}{2}mv^2$$

C
$$qvBr$$

D
$$mv^2$$

E
Cannot be determined

Solution

As the charge is moving perpendicular to the magnetic field, therefore, direction of force on charge by magnetic field will be given by Fleming's left hand rule, that is vertically downward (perpendicular to both velocity and magnetic field).
now work done is given by,
$$W=Fd\cos\theta$$ ,
but $$\theta=90^0$$, as force and displacement (velocity) are perpendicular to each other,
therefore, $$W=Fd\cos90^0=0$$
Question 4
1 / -0

Two wires carrying current in opposite directions are placed a distance $$a$$ apart.
Which of the following will cause the greatest increase in the magnitude of the force between the wires?

A
Doubling the current in one wire.

B
Doubling the current in both wires.

C
Doubling the distance between the wires to $$2a$$.

D
Decreasing the distance between the wires to $$\frac{a}{2}$$

E
Running the currents in the same direction.

Solution

In this case, the force per unit length between the wires is $$F=\dfrac{\mu_0I_1I_2}{2\pi a}$$
As force is depends on current in both wires so when we double the current in both wire we get greatest increase of force $$F$$.
Thus, option B will be correct.
Question 5
1 / -0

Three wires $$A,B$$, and $$C$$ all are all parallel and have the same magnitude of current. Wires $$A$$ and $$B$$ have current going toward the top of the screen while wire $$C$$ has current going toward the bottom of the screen.
What is the direction of the force on $$A$$ because of the other two wires?

A
To the right.

B
To the left.

C
Into the screen.

D
Out of the screen.

Solution

From Fleming's Right Hand thumb rule, the magnetic field acting on wire A due to wire B is out of the page, whereas due to C is into the page. Now the magnetic field due to a current carrying wire decreases with increasing distance from the wire. Hence magnetic field at A due to C is weaker than that due to B. Therefore net magnetic field acts out of the page at A.
From $$\vec{F}=I(\vec{l}\times \vec{B})$$, the net force acts to the right.
Question 6
1 / -0

Two wires carrying equal current $$I$$ are a distance $$a$$ apart.
If the current through wire one is doubled while the current through the wire two is tripled, what is the ratio of the force that wire two exerts on wire 1 ($$F_1$$) to the force that wire one exerts on wire 2 ($$F_2$$)?

A
4:09

B
2:03

C
1:01

D
3:02

E
9:04

Solution

The force between two long parallel current ($$I_{1} , I_{2}$$) carrying wire separated by a distance $$a$$ is given by ,
$$F=\dfrac{\mu_{0}I_{1}I_{2}L}{2\pi a}$$ ,
this is the force exerted by wire 2 on wire 1 and by wire 1 on wire 2 , which depends on the product of current in both wire , therefore change in currents in the wire will affect the force between them but they will exert equal force on each other i.e. $$F_{1}/F_{2}=1/1$$ , or $$F_{1}:F_{2}=1:01$$
Question 7
1 / -0

Two wires carrying equal current $$I$$ in opposite directions are placed a distance $$a$$ apart. The wires experience a force $$F$$ due to the current. The wires are moved to a distance $$\dfrac{a}{2}$$.
If the current in wire one remains the same, which of the following values of $$I_2$$ would result in the same force $$F$$ between the wires?

A
$$\dfrac{1}{4}I$$

B
$$\dfrac{1}{2}I$$

C
$$I$$

D
$$2I$$

E
$$4I$$

Solution

The force acting on the wires initially $$F = \dfrac{\mu_o I_1I_2}{2\pi a}$$
where $$I_1 = I_2 = I$$
$$\implies$$ $$F = \dfrac{\mu_o I^2}{2\pi a}$$
New distance between the wires is $$a'$$ with current in the wires $$I_1' $$ & $$I_2'$$.
Given : $$a ' = \dfrac{a}{2}$$ and $$I_1' = I$$
Thus new force $$F' = \dfrac{\mu_o I I_2'}{2\pi (a/2)} = \dfrac{\mu_o II_2'}{\pi a}$$
But $$F' = F$$
$$\implies I_2' = \dfrac{I}{2}$$
Question 8
1 / -0

Two current carrying wires are a distance $$a$$ apart and experience a Force $$F$$ between them. The wires are moved a distance 2$$a$$ apart.
Which of the following is a possible value for the new force between the wires?

A
$$\dfrac{1}{4}F$$

B
$$\dfrac{1}{2}F$$

C
$$F$$

D
$$2F$$

E
$$4F$$

Solution

Force between two parallel current carrying wires $$F = \dfrac{\mu_o}{2\pi} \dfrac{I_1I_2}{a}$$
New distance between the wires $$a' =2a$$
$$\therefore$$ New force $$F' = \dfrac{\mu_o}{2\pi}\dfrac{I_1I_2}{2a} = \dfrac{F}{2}$$
Question 9
1 / -0

Two wires are pictured below, both carrying current toward the east.
What is the direction of the force exerted by wire $$2$$ on wire $$1$$?

A
North

B
South

C
Up

D
Down

E
East

Solution

Using right hand thumb rule, magnetic field at point P due to wire 2 $$\vec{B} = B\hat{z}$$
From figure, $$\vec{l} = l$$ $$\hat{x}$$
Magnetic force acting on wire 1, $$\vec{F} = I (\vec{l} \times \vec{B})$$
$$\therefore$$ $$\vec{F} = I(l\hat{x} \times B\hat{z}) = -IlB$$ $$\hat{y}$$
Thus magnetic force on wire 1 acts in south direction.
Question 10
1 / -0

A wire carrying a current of 4 A is in a 0.5 T magnetic field as shown above.
If the direction of electron flow is to the right as shown, what is the magnitude and direction of the force (per unit length) on the wire?

A
$$2N/m$$ into the page

B
$$8N/m$$ out of the page

C
$$2N/m$$ to the top of the page

D
$$2N/m$$ to the bottom of the page

E
$$8N/m$$ to the top of the page

Solution

Given : $$\vec{B} = -0.5$$ $$\hat{z}$$
Current flowing through the wire $$\vec{I} = - 4 \hat{x}$$ A (because current flows in a direction opposite to that of electron flow)
Let the length of the wire be $$l$$.
Magnetic force $$\vec{F} = I(\vec{l} \times \vec{B}) = I lB[-\vec{x} \times (-\hat{z})]$$
$$\therefore$$ Force per unit length $$\dfrac{\vec{F}}{l} =( 4 \times 0.5) $$ $$(-\hat{y}) =-2 $$ $$\hat{y}$$ $$N/m$$
Hence option D is correct.