Moving Charges and Magnetism Test

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Moving Charges and Magnetism Test - 81

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Question 1
1 / -0

A $$25cm$$ long solenoid has radius $$2cm$$ and $$500$$ total number of turns. It carries a current of $$15A$$. If it is equivalent to a magnet of the same size and magnetization $$\overrightarrow { M } $$ (magnetic moment/volume), then $$\left| \overrightarrow { M } \right| $$ is:

A
$$3000\pi A{m}^{-1}$$

B
$$3\pi A$$ $${m}^{-1}$$

C
$$30000 A$$ $${m}^{-1}$$

D
$$300 A$$ $${m}^{-1}$$

Solution
Question 2
1 / -0

Two parallel wires in free space are $$10cm$$ apart and cach carries a current of 10 $$ \mathrm{A} $$ in the same direction. The magnetic force per unit length of cach wire is.

A
$$ 2 \times 10^{-4} \mathrm{N}, $$ attractive

B
$$ 2 \times 10^{-4} \mathrm{N}, $$ repulsive

C
$$ 2 \times 10^{-7} \mathrm{N}, $$ attractive

D
$$ 2 \times 10^{-7} \mathrm{N}, $$ repulsive

Solution
Question 3
1 / -0

A particle having the same charge as of electron moves in a circular path of radius 0.5 cm under the influence of a magnetic field of 0.5 T. If an electric field of 100 V/m makes it move in a straight path, then the mass of the particle is ___?
(Given charge of electron = $$1.6 \times 10^{-19} C$$)

A
$$2.0 \times 10^{-24} kg$$

B
$$1.6 \times 10^{-19} kg$$

C
$$1.6 \times 10^{-27} kg$$

D
$$9.1 \times 10^{-31} kg$$

Solution

$$\dfrac{mv^2}{R} = qvB$$
$$mv = q BR$$
Path is straight line
it qE = qvB
E = vB
From equation (i) & (ii)
$$m = \dfrac{qB^2R}{E}$$
$$m = 2.0 \times 10^{-24} kg$$
Question 4
1 / -0

An electron moves with velocity $$v$$ in uniform traverse magnetic field $$B$$ on circular path of radius $$'r'$$, then $$e/m$$ for it is

A
$$\frac{v}{{Br}}$$

B
$$\frac{B}{{rv}}$$

C
$$Bvr$$

D
$$\frac{{vr}}{B}$$

Solution
Question 5
1 / -0

A particle enters the region of a uniform magnetic field as shown in figure. The path of the particle inside the field is shown by dark line .
The particle is :

A
electrically neutral

B
positively charged

C
negative charged

D
information given is adequate

Solution
Question 6
1 / -0

Which of following cannot be deflected by magnetic field

A
$$\alpha -$$ rays

B
$$\beta$$ rays

C
$$\gamma -$$ rays

D
cosmic rays

Solution

$$ \gamma- rays$$ will not be deflected by the magnetic field because $$\gamma- rays$$ don't have any charge.
Question 7
1 / -0

Two parallel infinite wires separated by distance $$'d'$$ carry currents as shown in figure.
The distance from a third infinite wire be kept parallel to wire carrying current $$l_1$$, the wire such that it stays in equilibrium is

A
$$\dfrac{I_2}{I_2 + I_1}d$$ or $$\dfrac{I_1}{I_1 + I_2}d$$

B
$$\dfrac{I_2}{I_2 - I_1}d$$ or $$\dfrac{I_1}{I_1 - I_2}d$$

C
$$\dfrac{I_2}{I_1 - I_2}d$$ or $$\dfrac{I_1}{I_1 - I_2}d$$

D
$$\dfrac{2I_2}{I_2 + I_1}d$$ or $$\dfrac{I_1}{I_1 - I_2}d$$

Solution

For the case when $$I_1 < I_2$$
Let the length of the third wire is $$\ell (\ell \rightarrow \infty)$$
For equilibrium $$F_1 = F_2$$
$$\Rightarrow \dfrac{\mu_0 I_1I}{2 \pi x} \ell = \dfrac{\mu_0 I_2I}{2 \pi (d + x)} \ell \Rightarrow \dfrac{d + x}{x} = \dfrac{I_2}{I_1} \Rightarrow \dfrac{d}{x} = \dfrac{I_2 - I_1}{I_2} \Rightarrow x = \left(\dfrac{I_2}{I_2 - I_1} \right) d$$
For the case when $$I_2 < I_1$$
$$F_1 = F_2$$
$$\dfrac{\mu_0 I_1 I \ell}{2 \pi x} = \dfrac{\mu_0 I_2 I \ell}{2 \pi (x - d)}$$
$$\Rightarrow \dfrac{x - d}{x} = \dfrac{I_2}{I_1}$$
$$\Rightarrow 1 - \dfrac{d}{x} = \dfrac{I_2}{I_1} \Rightarrow \dfrac{I_1 - I_2}{I_1} = \dfrac{d}{x}$$
$$\Rightarrow x = \left(\dfrac{I_1}{I_1 - I_2} \right) d$$
$$\therefore$$ value of x is $$\dfrac{I_2}{I_2 - I_1}$$ or $$\dfrac{I_1}{I_1 - I_2} d$$
Question 8
1 / -0

A moving coil galvanometer, having a resistance $$G$$, produces full scale deflection when a current $$I_g$$ flows through it. This galvanometer can be converted into (i) an ammeter of range $$0$$ to $$I_0 (I_0 > I_g)$$ by connecting a shunt resistance $$R_A$$ to it and (ii) into a voltmeter of range $$0$$ to $$V(V = GI_0)$$ by connecting a series resistance $$R_V$$ to it. Then

A
$$R_A R_V = G^2 \left(\dfrac{I_g}{I_0 - I_g}\right)$$ and $$\dfrac{R_A}{R_V} = \left(\dfrac{I_0 - I_g}{I_g}\right)$$

B
$$R_AR_V = G^2$$ and $$\dfrac{R_A}{R_V} = \left(\dfrac{I_g}{I_0 - I_g}\right)^2$$

C
$$R_AR_V = G^2$$ and $$\dfrac{R_A}{R_V} = \dfrac{I_g}{I_0 - I_g}$$

D
$$R_A R_V = G^2 \left(\dfrac{I_0 - I_g}{I_g}\right)$$ and $$\dfrac{R_A}{R_V} = \left(\dfrac{I_g}{I_0 - I_g}\right)^2$$

Solution

When galvanometer is used as an ammeter shunt is used in parallel with galvanometer.
$$\therefore I_g G = (I_0 - I_g)R_A$$

$$\therefore R_A = \left(\dfrac{I_g}{I_0 - I_g}\right) G$$

When galvanometer is used as a voltmeter, resistance is used in series with galvanometer.
$$I_g(G + R_v) = V = GI_0$$ (given $$V = GI_0$$)

$$\therefore R_v = \dfrac{(I_0 - I_g)G}{I_g}$$

$$\therefore R_AR_V = G^2 \& \dfrac{R_A}{R_V} = \left(\dfrac{I_g}{I_0 - I_g}\right)^2$$
Question 9
1 / -0

Using Kirchhoff's current law and Ohm's law. Find the magnitude of voltage V ( In volt) direction of current is as shown in figure

A
2V

B
3V

C
4V

D
5V
Question 10
1 / -0

A particle of charge -q and mass m enters a uniform magnetic field $$ \overrightarrow B $$ at A with speed $$ v_1 $$ at an angle $$ \alpha $$ and leaves the field at C with speed $$ v_2 $$ at an angle $$ \beta $$ as shown . then

A
$$ \alpha = \beta $$

B
$$ V_1 = V_2 $$

C
particle remains in the field for time $$ t = \dfrac {2 m ( \pi - \alpha)}{qB} $$

D
All of these

Solution

$$Given:$$ charge=$$-q$$ ; mass= $$m$$
$$Solution:$$
a) As the magnetic field is uniform, the charge comes out symmetrically. Hence, $$\boldsymbol{\alpha}=\boldsymbol{\beta}$$
b) $$V_1=V_{2}$$ since the speed does not change during motion in a plane perpendicular to the magnetic field.

c) $$\mathbf{T}=\frac{2 \pi m}{q B}$$
Time spent inside magnetic field is
$$\mathrm{T} \times \frac{(2 \pi-2 \alpha)}{2 \pi}=\frac{2 \pi m}{q B} \times \frac{(2 \pi-2 \alpha)}{2 \pi}=$$
$$\frac{2 m}{q B}(\pi-\alpha)$$
$$So,the $$ $$correct$$ $$option:D$$