Moving Charges and Magnetism Test

Result

Moving Charges and Magnetism Test - 79

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

A particle of charge per unit mass $$\alpha$$ is released from origin with a velocity $$\mathop v\limits^ \to = {v_0}\mathop i\limits^ \wedge $$ in a uniform magnetic field $$\mathop B\limits^ \to = - {B_0}\mathop k\limits^ \wedge $$ . If the particle passes through (0,y,0) then y is
equal to

A
$$ - \frac{{2{v_0}}}{{{B_0}\alpha }}\,\;$$

B
$$\frac{{{v_0}}}{{{B_0}\alpha }}\;$$

C
$$\;\frac{{2{v_0}}}{{{B_0}\alpha }}$$

D
$$ - \frac{{{v_0}}}{{{B_0}\alpha }}$$

Solution
Question 2
1 / -0

$$2$$ A current is flowing in a circular loop of radius $$1$$ m. Magnitude field at the centre of circular loop will be :-

A
$$\dfrac{\mu_0}{2}$$

B
$$2\mu_0$$

C
$$\mu_0$$

D
$$\dfrac{\mu_0}{2\pi}$$

Solution
Question 3
1 / -0

The magnitude of magnetic field due to current carrying arc of radius R, having a current l substanding an angle of $${60^0}$$ at the centre O is.

A
$$\frac{{{\mu _0}l}}{{8R}}$$

B
$$\frac{{{\mu _0}l}}{{10R}}$$

C
$$\frac{{2{\mu _0}l}}{{4\pi R}}$$

D
$$\frac{{{\mu _0}l}}{{12R}}$$

Solution
Question 4
1 / -0

iF $$\bar B_1,\bar B_2 and \bar B_3 $$ are the magnetic field due to $$i_1, I_2,and I_3$$ then in Ampere's circuital law $$\oint \bar B-d\bar I= \mu_0I, \bar B$$ is

A
$$\bar B=\bar B_1-\bar B_2$$

B
$$\bar B=\bar B1+\bar B_2+\bar B_3$$

C
$$\bar b=\bar B_1-\bar B_2+\bar B_3$$

D
$$\bar B=\bar B_3$$

Solution
Question 5
1 / -0

A particle of mass $$m$$ carrying charge $$q$$ is lying at the origin at the uniform magnetic field directed along $$Y-$$ axis. At time $$t=0$$, it is given velocity $$v_{0}$$ at angle $$\alpha $$ with the $$X-$$ axis as shown in the figure. The coordinates of the particle after one revolution is

A
$$(0, ({ v }_{ 0 }\cos { \alpha }) \cfrac { 2\pi m }{ qB }, 0)$$

B
$$(0, -({ v }_{ 0 }\cos { \alpha }) \cfrac { 2\pi m }{ qB }, 0)$$

C
$$(0, -({ v }_{ 0 }\sin { \alpha }) \cfrac { 2\pi m }{ qB }, 0)$$

D
$$(0, ({ v }_{ 0 }\sin { \alpha }) \cfrac { 2\pi m }{ qB }, 0)$$

Solution
Question 6
1 / -0

Two long wire each parallel to the $$z-$$axis and each having current $$i$$ are passing through origin and $$(x, y, 0)$$ respectively. The force per unit length on each wire is

A
$$\dfrac{\mu_0i^2}{2\pi(x^2+y^2)}$$

B
$$\dfrac{\mu_0 i^2}{2\pi\sqrt {x^2+y^2}}$$

C
$$\dfrac{\mu_0i^2}{2\pi(x+y)}$$

D
$$\dfrac{\mu_0i^2(x+y)}{2(x^2+y^2)}$$

Solution
Question 7
1 / -0

Rank the value of $$\oint{\overset{-}{B}}.dl$$ for the paths shown in figure from the smallest to largest:

A
a, b, c, d

B
a, c, d, b

C
a, d, c, b

D
a, c, b, d

Solution
Question 8
1 / -0

An electron at point A in figure has a speed $$v_0$$ of magnitude 1.41 x $$10^6$$ m/s. It enters into a uniform magnitude field and follows a semicircular path in it as shown.
The time required for the electron to move from A to B is

A
1.11 x $$10^{-7}$$ s

B
1.11 x $$10^6$$ s

C
1.11 x $$10^5$$ s

D
1.11 s

Solution
Question 9
1 / -0

In the figure, a charged sphere of mass $$m$$ and charge $$q$$ starts sliding from rest on a vertical fixed circular track of radius $$R$$ from the position shown. There exists a uniform and constant horizontal magnetic field induction $$B$$. The maximum force exerted by the track on the sphere is :

A
$$mg+qB \sqrt{2gR}$$

B
$$3mg-qB \sqrt{2gR}$$

C
$$3mg+qB\sqrt2gR$$

D
$$mg+qB\sqrt2gR$$

Solution

$${F_m} = qvB,$$ and directed radially outward$$.$$

$$\because N - mg\sin \theta + qvB = \dfrac{{m{v^2}}}{R}$$

$$ \Rightarrow N = \dfrac{{m{v^2}}}{R} + mg\sin \theta - qvB$$

Hence at $$\theta = \frac{\pi }{2}$$

$$ \Rightarrow {N_{\max }} = \dfrac{{2mgR}}{R} + mg - qB\sqrt {2gR} $$

$$ = 3mg - qB\sqrt {2gR} $$

Hence,
option $$(B)$$ is correct answer.
Question 10
1 / -0

In a toroid the number of turns per unit length is $$1000$$ and current through it is $$\dfrac{1}{4\pi}$$ ampere. The magnetic field produced inside(in weber/$$m^2$$) will be?

A
$$10^{-2}$$

B
$$10^{-3}$$

C
$$10^{-4}$$

D
$$10^{-7}$$

Solution