Moving Charges and Magnetism Test

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Moving Charges and Magnetism Test - 71

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Weekly Quiz Competition

Question 1
1 / -0

In Figure, assume $$I_1=2.00 \,A$$ and $$I_2=6.00 \,A$$. What is the relationship between the magnitude $$F_1$$ of the force exerted on wire $$1$$ and the magnitude $$F_2$$ of the force exerted on wire $$2$$?

A
$$F_1=6F_2$$

B
$$F_1=3F_2$$

C
$$F_1=F_2$$

D
$$F_1=\dfrac{1}{3}F_2$$

E
$$F_1=\dfrac{1}{6}F_2$$
Question 2
1 / -0

Classify each of the following statements as a characteristic.
The magnitude of the force depends on the charged object’s direction of motion.

A
of electric forces only

B
of magnetic forces only,

C
of both electric and magnetic forces, or

D
of neither electric nor magnetic forces.
Question 3
1 / -0

A proton moving horizontally enters a region where a uniform magnetic field is directed perpendicular to the protons velocity as shown in Figure. After the proton enters the field, does it.

A
deflect downward, with its speed remaining constant;

B
deflect upward, moving in a semicircular path with constant
speed, and exit the field moving to the left;

C
continue to move in the horizontal direction with constant velocity;

D
move in a circular orbit and become trapped by the field; or

E
deflect out of the plane of the paper?

Solution
Question 4
1 / -0

Classify each of the following statements as a characteristic.
The force exerted on a charged object is proportional to its speed.

A
of electric forces only

B
of magnetic forces only,

C
of both electric and magnetic forces, or

D
of neither electric nor magnetic forces.

Solution

$$F_B=|q|vB\sin\theta$$ is non-zero unless $$\theta=\pm90^{o}$$.
Question 5
1 / -0

In the velocity selector shown in Figure, electrons with speed $$v =\dfrac{ E}{B}$$ follow a straight path. Electrons moving significantly faster than this speed through the same selector will move along what kind of path?

A
a circle

B
a parabola

C
a straight line

D
a more complicated trajectory

Solution

The electrons will feel a constant electric force and a magnetic force that will charge in direction and in magnitude as their speed changes.
Question 6
1 / -0

Classify each of the following statements as a characteristic.
The force exerted on a moving charged object is zero.

A
of electric forces only

B
of magnetic forces only,

C
of both electric and magnetic forces, or

D
of neither electric nor magnetic forces.

Solution

But $$F_B=|q|vB\sin\theta$$ is zero if $$\theta=\pm90^{o}$$
Question 7
1 / -0

An electron is projected into a magnetic field of $$B = 5\times 10^{-3} T$$ and rotates in a circle of radius of $$R = 3\ mm$$. Find the work done by the force due to magnetic field.

A
$$0\ J$$

B
$$15\ mJ$$

C
$$14\ mJ$$

D
$$20\ mJ$$
Question 8
1 / -0

A current of i ampere is flowing through each ofthe bent wires as shown the magnitude and direction of magnetic field at O is

A
$$\displaystyle \frac{\mu_{0}\mathrm{i}}{4}(\frac{1}{\mathrm{R}}+\frac{2}{\mathrm{R}'})$$

B
$$\displaystyle \frac{\mu_{0}\mathrm{i}}{4}(\frac{1}{\mathrm{R}}+\frac{3}{\mathrm{R}'})$$

C
$$\displaystyle \frac{\mu_{0}\mathrm{i}}{8}(\frac{1}{\mathrm{R}}+\frac{3}{2\mathrm{R}}\prime)$$

D
$$\displaystyle \frac{\mu_{0}\mathrm{i}}{8}(\frac{1}{\mathrm{R}}+\frac{3}{\mathrm{R}'})$$

Solution

$$B=\dfrac { { \mu }_{ 0 }I }{ 2R } \dfrac { 3 }{ 4 } +\dfrac { { \mu }_{ 0 }I }{ 2R } \frac { 1 }{ 4 } $$

$$B=\dfrac { { \mu }_{ 0 }I }{ 8 } \left[ \dfrac { 3 }{ { R }^{ \prime } } +\dfrac { 1 }{ R } \right] $$
Question 9
1 / -0

When a particle of charge q is projected with uniform speed u along x - axis of the Cartesian coordinate system $$\left ( \hat{i},\hat{j},\hat{k} \right )$$ in the presence of a magnetic field of induction $$\vec{B}$$, the force on q is $$\vec{F}=\frac{qu\vec{B}}{2}\hat{j}$$ and when the particle is projected along y - axis with same speed, the force $$\vec{F}=\frac{qu\vec{B}}{2}\left ( -\hat{i}+\sqrt{3}\hat{k} \right )$$. The magnetic field $$\vec{B}$$ is

A
$$\frac{B}{2}(\sqrt{3}\hat{i}+\hat{k})$$

B
$$\frac{B}{2}(\sqrt{3}\hat{i}-\hat{k})$$

C
$$-\frac{B}{2}(\sqrt{3}\hat{i}+\hat{k})$$

D
$$\frac{B}{2}(-\sqrt{3}\hat{i}+\hat{k})$$

Solution

Initial velocity along x-axis $$U =U \hat{i}$$
Force when projected along x-axis $$F=q\left ( \vec{V}\times \vec{B} \right )$$
Let the magnetic field be $$\vec{B}=B\left ( a\hat{i}+b\hat{j}+c\hat{v} \right )$$
$$V_{1}=U\hat{i}$$
$$F_{1}=\frac{qUB}{2}\hat{j}=qB\left ( U\hat{i}\times \left ( a\hat{i}+b\hat{j} +c\hat{k}\right ) \right )$$
$$\frac{qUB}{2}\hat{j}=qBu\left ( v\hat{k}-c\hat{j} \right )$$
$$ b\hat{k}-c\hat{j}=\frac{\hat j}{2}$$
$$c=-\frac{1}{2},b=0 $$
$$F_{2}=\frac{qU\vec{B}}{2}\left ( -\hat{i} +\sqrt{3}\hat{k}\right ),V_{2}=U\hat{j}$$
$$F_{2}=q\left ( U\hat{j} \times \vec{B}\left ( a\hat{i}+c\hat{k} \right )\right )$$
$$\frac{qV\vec{B}}{2}\left ( -\hat{i}+\sqrt{3}\hat{k} \right )=qV\vec{B}\left ( -\hat{k} a+c\hat{i}\right )$$

$$a=-\frac{\sqrt{3}}{2},c=-\frac{1}{2}$$

$$\vec{B}=-\vec{B}\left ( \frac{\sqrt{3}}{2} \hat{i}+\frac{1}{2}\hat{k}\right )$$
Question 10
1 / -0

A charged particle q is moving with a velocity $$\vec{v_{1}}=2\hat{i}m/s$$ at a point in a magnetic field $$\vec{B}$$ and experiences a force $$\vec{F_{1}}=q(\hat{k}-2\hat{j})N.$$ If the same charge moves with velocity $$\vec{v_{2}}=2\hat{j}m/s$$ from the same point in that magnetic field and experiences a force $$\vec{F_{2}}=q(2\hat{i}+\hat{k})N$$, the magnetic induction at that point will be :

A
$$\hat{i}+\dfrac{1}{2}\hat{j}-\dfrac{1}{2}\hat{k}$$

B
$$-\dfrac{1}{2}\hat{i}+\dfrac{1}{2}\hat{j}+\hat{k}$$

C
$$\dfrac{1}{2}\hat{i}+\dfrac{1}{2}\hat{j}+\hat{k}$$

D
$$-\dfrac{1}{2}\hat{i}+\hat{j}+\dfrac{1}{2}\hat{k}$$

Solution

Let the magnetic induction at required point be $$\bar { B } ={ B }_{ x }\hat { i } +{ B }_{ y }\hat { j } +{ B }_{ z }\hat { k } $$
Magnetic force $$\left( \bar { F } \right) $$ on particle of charge (q), moving with velocity $$\bar { V } $$ in the magnetic field region of intensity $$\left( \bar { B } \right) $$ is given as
$$\bar { F } =q\left( \bar { V } \times \bar { B } \right) $$
Case 1: $$\bar { { F }_{ 1 } } =q\left( \bar { { V }_{ 1 } } \times \bar { B } \right) $$
$$q\left( \hat { k } -2\hat { j } \right) =q\left( 2\hat { i } \times \left( { B }_{ x }\hat { i } +{ B }_{ y }\hat { j } +{ B }_{ z }\hat { k } \right) \right) $$
$$\hat { k } -2\hat { j } =2{ B }_{ y }\hat { k } -2{ B }_{ z }\hat { j } $$
$$\therefore 2{ B }_{ y }=1,\quad { B }_{ y }={ 1 }/{ 2 }.....\left( i \right) $$
$$-2=-2{ B }_{ z },\therefore { B }_{ z }=1.....\left( ii \right) $$
Case 2: $$\bar { { F }_{ 2 } } =q\left( \bar { { V }_{ 2 } } \times \bar { B } \right) $$
$$q\left( 2\hat { i } +\hat { k } \right) =q\left( 2\hat { j } \times \left( { B }_{ x }\hat { i } +{ B }_{ y }\hat { j } +{ B }_{ z }\hat { k } \right) \right) $$
$$2\hat { i } +\hat { k } =-2{ B }_{ x }\hat { k } +2{ B }_{ z }\hat { i } $$
$$\therefore -2{ B }_{ x }=1,\quad \therefore +{ B }_{ x }={ 1 }/{ 2 }......\left( iii \right) $$
Thus from $$\left( i \right) ,\left( ii \right) \& \left( iii \right) $$
$$\bar { B } =-\cfrac { 1 }{ 2 } \hat { i } +\cfrac { 1 }{ 2 } \hat { j } +\hat { k } $$