Moving Charges and Magnetism Test

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Moving Charges and Magnetism Test - 73

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Question 1
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An electron moving with a speed $$u$$ along the positive $$x-$$axis at $$y=0$$ enters a region of uniform magnetic field which exists to the right of $$y-$$axis. The electron exits from the region after some time with the speed $$v$$ at coordinate $$y,$$ then :

A
$$v > u, y < 0$$

B
$$v = u, y > 0$$

C
$$v > u, y > 0$$

D
$$v = u, y < 0$$

Solution

$$\overrightarrow{v} =v\hat{i}$$
$$\overrightarrow{B} =u(-\hat{k})$$
$$\overrightarrow{F}_{mag} = -e(\overrightarrow{v} \times \overrightarrow{B} )=euB(\hat{i} \times \hat{k})=-euB\hat{j}$$

Thus the electron gets deflected towards $$-ve\quad y-$$axis.
Since no work is done during motion in a magnetic field, electron's speed will remain constant.
Question 2
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A charge particle of charge $$q$$ is moving with speed $$v$$ in a circle of radius $$R$$ as shown in figure. Then the magnetic field at a point on axis of circle at a distance $$x$$ from centre is :

A
$$\dfrac{\mu _{0}}{4\pi }\dfrac{qV}{R^{2}}$$

B
$$\dfrac{\mu _{0}}{4\pi }\dfrac{qV}{(R^{2}+x^{2})}$$

C
$$\dfrac{\mu _{0}}{4\pi }\dfrac{qV}{x^{2}}$$

D
$$\dfrac{\mu _{0}}{4\pi }\dfrac{qVR}{(R^{2}+x^{2})^{3/2}}$$

Solution

Charge $$q$$ moving in a circle gives rise to a current $$I=\dfrac{q}{\dfrac{2\pi R}{v}}=\dfrac{qv}{2\pi R}$$
The magnetic field on the axis at a distance $$x$$ from the center of a current carrying circular loop of radius $$R$$ is:
$$B_{axis}=\dfrac{\mu_0IR^2}{2(R^2 + x^2)^\frac{3}{2}}=\dfrac{\mu_0\dfrac{qv}{2\pi R}R^2}{2(R^2 + x^2)^\frac{3}{2}}=\dfrac{\mu_0qvR}{4\pi(R^2 + x^2)^\frac{3}{2}}$$
Question 3
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A charged particle moves inside a pipe, which is bent as shown in the figure. If $$R < \dfrac {mv}{qB}$$, then the force exerted by the pipe on charged particle at $$P$$ is (Neglect gravity) :

A
toward the center

B
away from the center

C
zero

D
none of these

Solution

Radius of the circular path in magnetic field $$R=\dfrac{mv}{qB}$$
At point $$P$$ if we neglect weight then the centripetal force is provided by the magnetic force which is towards the center.
Question 4
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A particle of charge per unit mass $$\alpha$$ is released from origin with velocity $$\vec v=v_0\hat i$$ in a magnetic field $$\vec B=-B_0\hat k$$ for $$x\leq \dfrac {\sqrt 3}{2}\dfrac {v_0}{B_0\alpha}$$ and $$\vec B=0$$ for $$x > \dfrac {\sqrt 3}{2}\dfrac {v_0}{B_0\alpha}$$. The x-coordinate of the particle at time $$t\left ( > \dfrac {\pi}{3B_0\alpha}\right )$$ would be

A
$$\displaystyle \frac {\sqrt 3}{2}\frac {v_0}{B_0\alpha}+\frac {\sqrt 3}{2}-v_0\left (t-\frac {\pi}{B_0\alpha}\right )$$

B
$$\displaystyle \frac {\sqrt 3}{2}\frac {v_0}{B_0\alpha}+v_0\left (t-\frac {\pi}{3B_0\alpha}\right )$$

C
$$\displaystyle \frac {\sqrt 3}{2}\frac {v_0}{B_0\alpha}+\frac {v_0}{2}\left (t-\frac {\pi}{3B_0\alpha}\right )$$

D
$$\displaystyle \frac {\sqrt 3}{2}\frac {v_0}{B_0\alpha}+\frac {v_0t}{2}$$

Solution

$$r=\dfrac {mv_0}{B_0q}=\dfrac {v_0}{B_0\alpha}, \dfrac {x}{r}=\dfrac {\sqrt 3}{2}=sin \theta$$

$$\Rightarrow \theta=60^o$$

$$t_{OA}=\dfrac {T}{6}=\dfrac {\pi}{3B_0\alpha}$$

Therefore, x-coordinate of particle at any time $$t > \dfrac {\pi}{3B_0\alpha}$$ will be
$$x=\dfrac {\sqrt 3}{2}\dfrac {v_0}{B_0\alpha}+v_0\left (t-\dfrac {\pi}{3B_0\pi}\right )cos 60^o$$

$$=\dfrac {\sqrt 3}{2}\dfrac {v_0}{B_0\alpha}+\dfrac {v_0}{2}\left (t-\dfrac {\pi}{3B_0\alpha}\right )$$
Question 5
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A charged particle of specific charge (charge/mass) $$\alpha$$ is released from origin at time $$t=0$$ with velocity $$\vec v=v_0(\hat i+\hat j)$$ in a uniform magnetic field $$\vec B=B_0\hat i$$. Coordinates of the particle at time $$t=\dfrac{\pi}{B_0\alpha}$$ are

A
$$\displaystyle\left (\frac {v_0}{2B_0\alpha}, \frac {\sqrt 2v_0}{\alpha B_0}, \frac {-v_0}{B_0\alpha}\right )$$

B
$$\displaystyle\left (\frac {-v_0}{2B_0\alpha}, 0, 0\right )$$

C
$$\displaystyle\left (0, \frac {2v_0}{B_0\alpha}, \frac {v_0\pi}{2B_0\alpha}\right )$$

D
$$\displaystyle\left (\frac {v_0\pi}{B_0\alpha}, 0, \frac {-2v_0}{B_0\alpha}\right )$$

Solution

$$\alpha=\dfrac {q}{m}$$, path of the particle will be a helix of time period,

$$T=\dfrac {2\pi m}{B_0q}=\dfrac {2\pi}{B_0\alpha}$$

The given time $$t=\dfrac {\pi}{B_0\alpha}=\dfrac {T}{2}$$

$$\therefore$$ Coordinates of particle at time $$t=T/2$$ would be $$(v_xT/2, 0, -2r)$$

Here, $$r=\dfrac {mv_0}{B_0q}=\dfrac {v_0}{B_0\alpha}$$

$$\therefore$$ The coordinate are $$\left (\dfrac {v_0\pi}{B_0\alpha}, 0, \dfrac {-2v_0}{B_0\alpha}\right )$$
Question 6
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Directions For Questions

The magnetic induction in vacuum at a plane surface of a magnetic material is equal to $$B$$ and the vector $$\vec{B}$$ forms an angle $$\theta$$ with the normal $$\vec{n}$$ of the surface (figure shown above). The permeability of the magnetic is equal to $$\mu$$.

...view full instructions

Find the circulation of the vector $$\vec{B}$$ around the square path $$T$$ with side $$l$$ located as shown in the figure above.

A
$$\displaystyle\oint{\vec{B}dr}=2(1-\mu)Bl\sin{\theta}$$

B
$$\displaystyle\oint{\vec{B}dr}=(1+\mu)Bl\sin{\theta}$$

C
$$\displaystyle\oint{\vec{B}dr}=2(1-+\mu)Bl\sin{\theta}$$

D
$$\displaystyle\oint{\vec{B}dr}=(1-\mu)Bl\sin{\theta}$$

Solution

magnetic field in the medium of magnetic permeability $$(\mu) $$ becomes $$\mu \times B_{vacuum}$$ .
$$\oint B.dr = B \sin\theta +[(- B\cos \theta) x + (-\mu B\cos\theta)(l-x)] + (-\mu B\sin\theta)l +[(B\cos\theta)(l-x) + (B\cos\theta)x]$$
$$\implies \oint B.dr = (1- \mu)Bl \sin\theta$$
Question 7
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A horizontal wire 0.1 m long carries a current of 5 A. Find the magnitude of the magnetic field, which can support the weight of the wire. Assume wire to be of mass $$3 \times 10^{-3}kg m^{-1}$$ :

A
$$5.88 \times 10^{-2}T$$

B
$$4.88 \times 10^{-3}T$$

C
$$5.88 \times 10^{-3}T$$

D
$$5.88 \times 10^{-4}T$$

Solution

Here, $$I=5 A; 1= 0.1 m$$;
Mass per unit length of the wire
$$\lambda =3 \times 10^{-3} kg m^{-1}$$
Therefore, weight of the wire,
$$mg=3 \times 10^{-3} \times 0.1 \times 9.8 = 2.94 \times 10^{-3}N$$
Let B be the strength of the magnetic field applied.
$$F=BIl = B \times 5 \times 0.1 = 0.5 B$$
In equilibrium,
$$F=mg$$
$$\therefore 0.5 B = 2.94 \times 10^{-3}$$
$$B \displaystyle =\frac{2.94 \times 10^{-3}}{0.5} = 5.88 \times 10^{-3}T$$
Question 8
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Two very long straight parallel wires carry steady currents $$i$$ and $$2i$$ in opposite directions. The distance between the wires is $$d$$. at a certain instant of time, a point charge $$q$$ is at a point equidistant from the two wires in the plane of the wires. Its instantaneous velocity $$\vec { v } $$ is perpendicular to this plane. The magnitude of the force due to the magnetic field acting on the charge at this instant is:

A
$$\cfrac { { \mu }_{ 0 }iqv }{ 2\pi d } $$

B
$$\cfrac { { \mu }_{ 0 }iqv }{ \pi d } $$

C
$$\cfrac { { 3\mu }_{ 0 }iqv }{ 2\pi d } $$

D
zero

Solution

Magnetic field at $$P$$ is perpendicu;ar to paper inward due to both the wires. Charged particle is also projected in the same direction.
So, force on the charged particle is zero as $$\vec { v } \parallel \vec { B

} ;\quad \vec { { F }_{ m } } =q(\vec { v } \times \vec { B } )=0$$
Question 9
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Directions For Questions

A charged particle carrying charge $$q=10 \mu C$$ moves with velocity $$v_1=10^6\ ms^{-1}$$ at angle $$45^o$$ with $$x-$$axis in the $$xy$$ plane and experiences of force $$F_1=5\sqrt 2\ mN$$ along the negative z-axis. When the same particle moves with velocity $$v_2=10^6\ ms^{-1}$$ along the $$z-$$axis, it experiences of force $$F_2$$ in $$y-$$direction.

...view full instructions

Find the magnetic field $$\vec B$$.

A
$$(10^{-3}T)(\hat i+\hat j)$$

B
$$(2\times 10^{-3}T)\hat i$$

C
$$(10^{-3}T)\hat i$$

D
$$(2\times 10^{-3}T)(\hat i_\hat j)$$

Solution

Force on a charged particle $$\vec{F} = q( \vec{v} \times \vec{B})$$

Let the magnetic field $$\vec{B} =B_x \hat{i} + B_y \hat{j} + B_z \hat{k}$$

$$1^{st}\ Case:$$

$$\vec{F} = (10 \times 10^{-6} \times 10^6)( \cos 45^{\circ}\hat{i} + \sin 45^{\circ} \hat{j}) \times (B_x \hat{i} + B_y \hat{j} + B_z \hat{k})$$

$$ \therefore 5 \times \sqrt{2} \times 10^{-3}(-\hat{k} )= \dfrac{10}{\sqrt{2}}(\hat{i} + \hat{j})\times (B_x \hat{i} + B_y \hat{j} + B_z \hat{k})= \dfrac{10}{\sqrt{2}}[B_z\hat{i} - B_z \hat{j} +(B_y -B_x)\hat{k} ] $$

Equating both LHS and RHS for components

$$B_z=0$$; $$B_y-B_x =-10^{-3} ..................................(1)$$

$$2^{nd}\ Case:$$

$$\vec{F} = (10 \times 10^{-6} \times 10^6)(\hat{k} ) \times (B_x \hat{i} + B_y \hat{j} + B_z \hat{k})$$

$$\therefore F_2 \hat{j} =(10 \times 10^{-6} \times 10^6)(B_x\hat{j} -B_y \hat{i})=10 \times(B_x\hat{j} -B_y \hat{i})$$

$$ \Rightarrow F_2 =10B_x$$; $$B_y=0 ...........................(2)$$

From equations $$(1)$$ and $$(2)$$

$$B_y-B_x =-10^{-3}= 0 - B_x =-10^{-3}$$

$$ \Rightarrow B_x= 10^{-3}\:T$$

$$ \Rightarrow \vec{B} = (10^{-3}\:T) \hat{i}$$
Question 10
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Directions For Questions

A charged particle carrying charge $$q=10 \mu C$$ moves with velocity $$v_1=10^6\ ms^{-1}$$ at angle $$45^o$$ with $$x-$$axis in the $$xy$$ plane and experiences of force $$F_1=5\sqrt 2\ mN$$ along the negative z-axis. When the same particle moves with velocity $$v_2=10^6\ ms^{-1}$$ along the $$z-$$axis, it experiences of force $$F_2$$ in $$y-$$direction.

...view full instructions

Find the magnitude of the force $$F_2$$

A
$$10^{-2}\ N$$

B
$$10^{-3}\ N$$

C
$$10^{-4}\ N$$

D
$$10^{-5}\ N$$

Solution

Force on a charged particle $$\vec{F} = q( \vec{v} \times \vec{B})$$

Let the magnetic field $$\vec{B} =B_x \hat{i} + B_y \hat{j} + B_z \hat{k}$$

$$1^{st}\ Case:$$

$$\vec{F}= (10 \times 10^{-6} \times 10^6)( \cos 45^{\circ}\hat{i} + \sin 45^{\circ} \hat{j}) \times (B_x \hat{i} + B_y \hat{j} + B_z \hat{k})$$

$$\therefore 5 \times \sqrt{2} \times 10^{-3}(-\hat{k} )= \dfrac{10}{\sqrt{2}}(\hat{i} + \hat{j})\times (B_x \hat{i} + B_y \hat{j} +B_z \hat{k})= \dfrac{10}{\sqrt{2}}[B_z\hat{i} - B_z \hat{j} +(B_y -B_x)\hat{k} ] $$

Equating both LHS and RHS for components

$$B_z=0$$; $$B_y-B_x =-10^{-3} .......................................(1)$$

$$2^{nd}\ Case:$$

$$\vec{F} = (10 \times 10^{-6} \times 10^6)(\hat{k} ) \times (B_x \hat{i} + B_y \hat{j} + B_z \hat{k})$$

$$\therefore F_2 \hat{j} =(10 \times 10^{-6} \times 10^6)(B_x\hat{j} -B_y \hat{i})=10 \times(B_x\hat{j} -B_y \hat{i})$$

$$ \Rightarrow F_2 =10B_x$$; $$B_y=0 .............................(2)$$

From $$(1)$$ and $$(2)$$

$$B_y-B_x =-10^{-3}= 0 - B_x =-10^{-3}$$

$$ \Rightarrow B_x= 10^{-3}\:T$$

$$ \Rightarrow \vec{B} = (10^{-3}\:T) \hat{i}$$

From eqn$$(2)$$

$$F_2= 10B_x= 10 \times 10^{-3}= 10^{-2}\:N$$