CBSE Questions for Class 12 Medical Physics Moving Charges And Magnetism Quiz 12 - MCQExams.com

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.
  • $$0\ J$$
  • $$15\ mJ$$
  • $$14\ mJ$$
  • $$20\ mJ$$
$$\mathrm{A}$$ magnetic field $$\vec{\mathrm{B}}=\mathrm{B}_{0}\hat{\mathrm{j}}$$ exists in the region a $$<\mathrm{x}<2\mathrm{a}$$ and $$\vec{\mathrm{B}}=-\mathrm{B}_{0}\hat{\mathrm{j}}$$ , in the region $$2\mathrm{a}<\mathrm{x}<3\mathrm{a}$$, where $$\mathrm{B}_{0}$$ is a positive constant. $$\mathrm{A}$$ positive point charge moving with a velocity $$\vec{\mathrm{v}}=\mathrm{v}_{0}\hat{\mathrm{i}}$$, where $$\mathrm{v}_{0}$$ is a positive constant, enters the magnetic field at $$\mathrm{x}=\mathrm{a}$$. The trajectory of the charge in this region can be like, 


42471.jpg

A bar magnet of magnetic moment $$M$$ is divided into '$$n$$' equal parts by cutting parallel to length. Then one part is suspended in a uniform magnetic field of strength $$2T$$ and held making an angle $$60^{0}$$ with the direction of the field. When the magnet is released, the kinetic energy of the magnet in the equilibrium position is:
  • $$\dfrac{M}{n}$$J
  • $$Mn$$ J
  • $$\dfrac{M}{n^{2}}$$J
  • $$Mn^{2}$$ J
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 :
  • $$\hat{i}+\dfrac{1}{2}\hat{j}-\dfrac{1}{2}\hat{k}$$
  • $$-\dfrac{1}{2}\hat{i}+\dfrac{1}{2}\hat{j}+\hat{k}$$
  • $$\dfrac{1}{2}\hat{i}+\dfrac{1}{2}\hat{j}+\hat{k}$$
  • $$-\dfrac{1}{2}\hat{i}+\hat{j}+\dfrac{1}{2}\hat{k}$$
The magnitude of force per unit length on a wire carrying current $$I$$ at O due to two semi infinite wires as shown in figure, (if the distance between the long parallel segments of the wire being equal to $$L$$ and current in them is $$I$$) is:

24049.png
  • $$\dfrac{+\mu_{0}I^{2}}{\pi L}\hat{j}$$
  • $$\dfrac{-\mu_{0}I^{2}}{L}\hat{j}$$
  • $$\dfrac{\mu_{0}I^{2}}{2L}\hat{j}$$
  • $$\dfrac{\mu_{0}I^{2}}{L}\hat{j}$$
The magnetic field due to a current carrying square loop of side a at a point located symmetrically at a distance of $$\mathrm{a}/2$$ from its centre (as shown in figure):

42292.jpg
  • $$\displaystyle \dfrac{\sqrt{2}\mu_{0}\mathrm{i}}{\sqrt{3}\pi \mathrm{a}}$$
  • $$\displaystyle \dfrac{\mu_{0}\mathrm{i}}{\sqrt{6}\pi \mathrm{a}}$$
  • $$\displaystyle \dfrac{2\mu_{0}\mathrm{i}}{\sqrt{3}\pi \mathrm{a}}$$
  • $$zero$$
A planar coil of area 7 $$\mathrm{m}^{2}$$ carrying an anti-clockwise current 2 A is placed in an extemal magnetic field $$\vec{B}=(0.2\hat{i}+0.2\hat{j}-0.3\hat{k})$$, such that the normal to the plane is along the line$$(3\hat{i}-5\hat{j}+4\hat{k})$$. Select correct statements from the following .  ( Consider Normal of the coil and Magnetic moment vectors to be in  the same direction ) 
  • The potential energy of the coil in the given orientation is $$6.4 J$$
  • The angle between the normal (positive) to the coil and the external magnetic field is $$\cos^{-1}\ (0.57)$$
  • The potential energy of the coil in the given orientatlon is $$3.2 J$$
  • The magnitude of magnetic moment of the coil is about 14 $$A{m}^{2}$$
A charged particle having charge q experiences a force $$\vec{F}=q(-\vec{j}+\vec{k})N$$ in a magnetic field B when it has a velocity $$v_{1} = 1\hat i \ m/s$$. The force becomes $$\vec{F}=q(\vec{i}-\vec{k})N$$ when the velocity is changed to $$v_{2}=1\hat{j}m/sec$$. The magnetic induction vector at that point is :
  • $$(\hat{i}+\hat{j}+\hat{k})\ T$$
  • $$(\hat{i}-\hat{j}-\hat{k})\ T$$
  • $$(-\hat{i}-\hat{j}+\hat{k})\ T$$
  • $$(\hat{i}+\hat{j}-\hat{k})\ T$$
A charged particle A of charge q = 2 C has velocity v = 100 m/s. When it passes through point A and has  velocity in the direction shown. The strength of magnetic field at point B due to this moving charge is (r = 2 m).the agle between them is 30.

42293_a3c6da88eb52467ead7e1a4036060562.png
  • 2.5$$\mu$$T
  • 5.0$$\mu$$T
  • 2.0$$\mu$$T
  • None
An $$\alpha $$ particle is moving along a circle of radius $$R$$ with a constant angular velocity $$\omega.$$ Point $$A$$ lies in the same plane at a distance $$2R$$ from the centre. Point $$A$$ records magnetic field produced by $$\alpha $$ particle. If the minimum time interval between two successive times at which $$A$$ records zero magnetic field is $$'t'$$, the angular speed $$\omega ,$$ in terms of $$t$$ is 
  • $$\displaystyle \frac{2\pi }{t}$$
  • $$\displaystyle \frac{2\pi }{3t}$$
  • $$\displaystyle \frac{\pi }{3t}$$
  • $$\displaystyle \frac{\pi }{t}$$

An electric motor is a device to convert electrical energy into mechanical energy. The motor shown below has a rectangular coil $$(15\ cm \times 10\ cm)$$ with 100 turns placed in a uniform magnetic fleld $$B=2.5\ T$$. When a current is passed through the coil, lt completes $$50$$ revolutions in one second. the power output of the motor is $$1.5\ kW$$. The current rating of the coil should be



44469.jpg
  • 1 A
  • 2 A
  • 3 A
  • 4 A
A thin rod of mass $$m$$ and length $$l$$ has charge $$Q$$, distributed uniformly along its length. It is rotated with angular speed $$\omega $$ in horizontal plane about an axis, passing through the mid-point and perpendicular to its length. If in the presence of an external magnetic field $$B_{o}$$ oriented along the axis of rotation, the total kinetic energy of the rod is reduced to zero, then the angular speed of the rod must be :
  • $$\dfrac{QB_{o}}{2m}$$
  • $$\dfrac{2QB_{o}}{m}$$
  • $$\dfrac{QB_{o}}{m}$$
  • $$\dfrac{QB_{o}}{4m}$$
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
  • $$\frac{B}{2}(\sqrt{3}\hat{i}+\hat{k})$$
  • $$\frac{B}{2}(\sqrt{3}\hat{i}-\hat{k})$$
  • $$-\frac{B}{2}(\sqrt{3}\hat{i}+\hat{k})$$
  • $$\frac{B}{2}(-\sqrt{3}\hat{i}+\hat{k})$$
A ring or radius $$R$$ moves with a velocity $$v$$ in a unifrom static magnetic field $$B$$ as shown in diagram. The emf between $$P$$ and $$Q$$ is
1092989_3f79b37a22084b789e382c4aca89521c.png
  • $$zero$$
  • $$vBR(1-\cos \theta)$$
  • $$2vBR\sin \dfrac {\theta}{2}$$
  • $$vB \sin\theta$$
A charged particle $$A$$ of charge $$q = 2\ C$$ has velocity $$v = 100 \ m/s$$. When it passes through point $$A$$ and has velocity in the direction shown, the strength of magnetic field at point $$B$$ due to this moving charge is $$(r = 2\ m)$$

76343.png
  • $$ 2.5\ \mu T$$ 
  • $$ 5.0\ \mu T $$
  • $$ 2.0\ \mu T$$
  • $$none\ of\ these$$
A charged particle moves through a magnetic field in a direction perpendicular to it. Then the
  • Speed of the particle remains unchanged
  • Direction of the particle remains unchanged
  • Acceleration remains unchanged
  • Velocity remains unchanged
A positive charge particle with velocity $$\overrightarrow{v} = x\hat{i} + y\hat{j}$$ moves in a magnetic field $$\overrightarrow{B} = y\hat{i} + x\hat{j}$$. The magnitude of magnetic force acting on the particle is F. Which one of the following statements are correct ?
  • no force will act on particle if $$x = y$$
  • $$F \propto (x^2 - y^2)$$ if $$x > y$$
  • the force will act along positive Z-axis if $$x > y$$
  • the force will act along positive Z-axis if $$y > x$$
A Positive charge particle with velocity $$\vec v=x\hat i+y\hat j$$ moves in a magnetic field $$\vec B=y\hat i+x\hat j$$. The magnitude of magnetic force acting on the particles is F. Which one of the following statements are correct :
  • no force will act on particle of x = y
  • $$F\propto (x^2-y^2)$$ if x > y
  • the force will act along positive Z-axis if x > y
  • the force will act along positive Z-axis if y > x
The field created by the current in the loop at point C will be
74380.png
  • $$-\dfrac{\mu _{0} }{4\pi }\hat{k}$$
  • $$-\dfrac{\mu _{0} }{8\pi }\hat{k}$$
  • $$-\dfrac{\mu _{0}\sqrt{2} }{\pi }\hat{k}$$
  • none
A charged particle is kept at rest in a uniform magnetic field. If the magnetic field increases with time, 
  • the particle starts moving
  • the particle undergoes circular motion
  • the particle starts moving in opposite direction
  • None of the above
The radius of the curved part of the wire is $$R=100\:mm$$, the linear parts of the wire are very long. Find the magnetic induction at the point $$O$$ if the wire carrying a current $$I=8.0\:A$$ has the shape shown in figure.
144843_c9ba9b4452f549f290ae745fc042154f.png
  • $$\displaystyle 0.34\:\mu T$$
  • $$\displaystyle 0.11\:\mu T$$
  • $$\displaystyle 1.1\:\mu T$$
  • $$\displaystyle 34\:\mu T$$
The radius of the curved part of the wire is $$R=100\:mm$$, the linear parts of the wire are very long. Find the magnetic induction at the point $$O$$ if the wire carrying a current $$I=8.0\ A$$ has the shape shown in figure.
144841_2fd978e7d32944af8e2645d0291e330b.png
  • $$0.44\:\mu T$$
  • $$0.34\:\mu T$$
  • $$0.56\:\mu T$$
  • $$0.78\:\mu T$$
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 :

77198.jpg
  • $$\dfrac{\mu _{0}}{4\pi }\dfrac{qV}{R^{2}}$$
  • $$\dfrac{\mu _{0}}{4\pi }\dfrac{qV}{(R^{2}+x^{2})}$$
  • $$\dfrac{\mu _{0}}{4\pi }\dfrac{qV}{x^{2}}$$
  • $$\dfrac{\mu _{0}}{4\pi }\dfrac{qVR}{(R^{2}+x^{2})^{3/2}}$$
The potential energy for a force field $$\vec {F}$$ is given by $$U(x,y)=\sin(x+y)$$. The force acting on the particle of mass $$m$$ at $$\left(0,\dfrac {\pi}{4}\right)$$ is
  • $$1$$
  • $$\sqrt {2}$$
  • $$\dfrac {1}{\sqrt {2}}$$
  • $$0$$
A non conductive insulating ring of mass $$m$$ and radius $$R$$, having charge $$Q$$ uniformly distributed over it's circumference is hanging by a insulated thread with the help of a small smooth ring (not rigidly fixed with bigger ring). A time varying magnetic field $$B=B_0 \sin \omega t$$ is switched at $$t=0$$ and ring is released at the same time. The average magnetic moment of Ring in time interval $$ 0 $$ is:

77083.jpg
  • $$\dfrac {B_0q^2R^2}{2\pi m}$$
  • $$\dfrac {B_0q^2R^2}{4\pi m}$$
  • $$\dfrac {B_0q^2R^2}{\pi m}$$
  • zero
The magnetic field $$B$$ at the centre of a circular coil of radius $$r$$ is $$\pi$$ times that due to a long straight wire at a distance $$r$$ from it, for equal currents. The following diagram shows three cases. In all cases the circular part has radius r and straight ones are infinitely long. For the same current the field $$B$$ at centre $$P$$ in cases $$1, 2, 3$$ has the ratio :

143606_ce1fcb87563e476bb841190c96ae32e8.png
  • $$\left (-\dfrac {\pi}{2}\right ):\left (\dfrac {\pi}{2}\right ):\left (\dfrac {3\pi}{4}-\dfrac {1}{2}\right )$$
  • $$\left (-\dfrac {\pi}{2}+1\right ):\left (\dfrac {\pi}{2}+1\right ):\left (\dfrac {3\pi}{4}+\dfrac {1}{2}\right )$$
  • $$\left (-\dfrac {\pi}{2}\right ):\left (\dfrac {\pi}{2}\right ):\left (\dfrac {3\pi}{4}\right )$$
  • $$\left (-\dfrac {\pi}{2}-1\right ):\left (\dfrac {\pi}{2}-\dfrac {1}{4}\right ):\left (\dfrac {3\pi}{4}+\frac {1}{2}\right )$$
L is a circular ring made of a uniform wire. Current enters and leaves the rind through straight conductors which, if produced, would have passed through the centre C of the ring. The magnetic field at C :

143090_8c9cabc4cab2406db41eb988d3ac0649.png
  • due to the straight conductors is zero
  • due to the loop is zero
  • due to the loop is proportional to $$\theta$$
  • due to the loop is proportional to $$(\pi -\theta)$$
Find the circulation of the vector $$\vec{B}$$ around the square path $$T$$ with side $$l$$ located as shown in the figure above.
145600_4b1ec8fc55ac4c5bb1af5841762e1767.png
  • $$\displaystyle\oint{\vec{B}dr}=2(1-\mu)Bl\sin{\theta}$$
  • $$\displaystyle\oint{\vec{B}dr}=(1+\mu)Bl\sin{\theta}$$
  • $$\displaystyle\oint{\vec{B}dr}=2(1-+\mu)Bl\sin{\theta}$$
  • $$\displaystyle\oint{\vec{B}dr}=(1-\mu)Bl\sin{\theta}$$
The radius of the curved part of the wire is $$R=100\:mm$$, the linear parts of the wire are very long. Find the magnetic induction at the point $$O$$ if the wire carrying a current $$I=8.0\ A$$ has the shape shown in the figure.
144839_a71ac60f4c2347d583788039a1fef47d.png
  • $$\displaystyle 0.30\:\mu T$$
  • $$\displaystyle 0.60\:\mu T$$
  • $$\displaystyle 0\:\mu T$$
  • $$\displaystyle 0.70\:\mu T$$
Velocity and acceleration vector of a charged particle moving in a magnetic field at some instant are $$\overrightarrow{v}=3\hat i+4\hat j$$ and $$\overrightarrow{a}=2\hat i+x\hat j$$. Select the correct options.
  • $$x=-1.5$$
  • $$x=3$$
  • Magnetic field has a component along the z-direction
  • Kinetic energy of the particle is constant
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
  • $$\displaystyle \frac {\sqrt 3}{2}\frac {v_0}{B_0\alpha}+\frac {\sqrt 3}{2}-v_0\left (t-\frac {\pi}{B_0\alpha}\right )$$
  • $$\displaystyle \frac {\sqrt 3}{2}\frac {v_0}{B_0\alpha}+v_0\left (t-\frac {\pi}{3B_0\alpha}\right )$$
  • $$\displaystyle \frac {\sqrt 3}{2}\frac {v_0}{B_0\alpha}+\frac {v_0}{2}\left (t-\frac {\pi}{3B_0\alpha}\right )$$
  • $$\displaystyle \frac {\sqrt 3}{2}\frac {v_0}{B_0\alpha}+\frac {v_0t}{2}$$
A particle having a mass of 0.5 g carries a charge of $$2.5\times 10^{-8}C$$. The particle is given an initial horizontal velocity of $$6\times 10^4 ms^{-1}$$. To keep the particle moving in a horizontal direction
  • the magnetic field may be perpendicular to the direction of the velocity
  • the magnetic field should be along the direction of the velocity
  • magnetic field should have a minimum value of 3.27 T
  • no magnetic field is required
An insulating rod of length $$l$$ carries a charge $$q$$ distributed uniformly on it. The rod is pivoted at its mid-point and is rotated at a frequency $$f$$ about a fixed axis perpendicular to the rod and passing through the pivot. The magnetic moment of the rod system is
  • $$\dfrac {1}{12}\pi q f l^2$$
  • $$\pi qfl^2$$
  • $$\dfrac {1}{6}\pi q f l^2$$
  • $$\dfrac {1}{3}\pi q f l^2$$
A charged particle $$P$$ leaves the origin with speed $$v=v_0$$ at some inclination with the $$x-$$axis. There is a uniform magnetic field B along the $$x-$$axis. $$P$$ strikes a fixed target $$T$$ on the $$x-$$axis for a minimum value of $$B=B_0$$. P will also strike T if
  • $$B=2B_0, v=2v_0$$
  • $$B=2B_0, v=v_0$$
  • $$B=B_0, v=2v_0$$
  • $$B=\dfrac {B_0}{2}, v=2v_0$$
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
  • $$\displaystyle\left (\frac {v_0}{2B_0\alpha}, \frac {\sqrt 2v_0}{\alpha B_0}, \frac {-v_0}{B_0\alpha}\right )$$
  • $$\displaystyle\left (\frac {-v_0}{2B_0\alpha}, 0, 0\right )$$
  • $$\displaystyle\left (0, \frac {2v_0}{B_0\alpha}, \frac {v_0\pi}{2B_0\alpha}\right )$$
  • $$\displaystyle\left (\frac {v_0\pi}{B_0\alpha}, 0, \frac {-2v_0}{B_0\alpha}\right )$$
A long circular tube of length $$10\ m$$ and radius $$0.3\ m$$ carries a current $$I$$ along its curved surface as shown. A wire-loop of resistance $$0.005\ ohm$$ and of radius $$0.1\ m$$ is placed inside the tube with its axis coinciding with the axis of the tube. The current varies as $$I = I_0\cos (300  t) $$ where $$I_0$$ is constant, If the magnetic moment of the loop is $$N \mu_0 I_0\sin (300  t)$$ then '$$N$$' is
166670.png
  • 6
  • 8
  • 7
  • 4
An electron is moving along the positive x-axis. You want to apply a magnetic field for a short time so that the electron may reverse its direction and move parallel to the negative x-axis. This can be done by applying the magnetic field along :
  • $$y-$$axis
  • $$z-$$axis
  • $$y-$$axis only
  • $$z-$$axis only
A charged particle with velocity $$\overrightarrow{v}=x\hat i+y\hat j$$ moves in a magnetic field $$\overrightarrow{B}=y\hat i+x\hat j$$. The force acting on the particle has magnitude $$F$$. Which one of the following statements is/are correct?
  • No force will act on charged particle if $$x=y$$
  • If $$x > y, F\propto(x^2-y^2)$$
  • If $$x > y$$, the force will act along z-axis
  • If $$y > x$$, the force will act along y-axis
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:
  • $$\cfrac { { \mu }_{ 0 }iqv }{ 2\pi d } $$
  • $$\cfrac { { \mu }_{ 0 }iqv }{ \pi d } $$
  • $$\cfrac { { 3\mu }_{ 0 }iqv }{ 2\pi d } $$
  • zero
Find the magnetic field $$\vec B$$.
  • $$(10^{-3}T)(\hat i+\hat j)$$
  • $$(2\times 10^{-3}T)\hat i$$
  • $$(10^{-3}T)\hat i$$
  • $$(2\times 10^{-3}T)(\hat i_\hat j)$$
Find the magnitude of the force $$F_2$$
  • $$10^{-2}\ N$$
  • $$10^{-3}\ N$$
  • $$10^{-4}\ N$$
  • $$10^{-5}\ N$$

The field B at the centre of a circular coil of radius $$r$$ is $$\pi$$ times that due to a long straight wire at a distance $$r$$ from it for equal currents. The figure shows three cases. In all cases the circular part has radius $$r$$ and straight ones are infinitely long. For same current, the field B at the centre P in cases 1, 2, 3 has the ratio:
166932.JPG
  • $$\displaystyle \left ( - \frac{\pi}{2} \right ) : \frac{\pi}{2} : \left ( \frac{3 \pi}{4} - \frac{1}{2} \right )$$
  • $$\displaystyle \left ( - \frac{\pi}{2} + 1\right ) : \left ( \frac{\pi}{2}+ 1 \right ) : \left ( \frac{3 \pi}{4} - \frac{1}{2} \right )$$
  • $$\displaystyle -\frac{\pi}{2} : \frac{\pi }{2} : \frac{3 \pi}{4}$$
  • $$\displaystyle \left ( - \frac{\pi}{2} - 1\right ) : \left ( \frac{\pi}{4}+ \frac{1}{4} \right ) : \left ( \frac{3 \pi}{4} + \frac{1}{2} \right )$$
A wire is bent in the form of a circular arc with a straight portion $$AB$$. Magnetic induction at $$O$$ when current $$I$$ is flowing in the wire, is

166713.PNG
  • $$\cfrac { { { \mu }_{ 0 } } }{ 2r } \left( \pi -\theta +\tan { \theta } \right) $$
  • $$\cfrac { { { \mu }_{ 0 } I} }{ 2\pi r } \left( \pi +\theta -\tan { \theta } \right) $$
  • $$\cfrac { { { \mu }_{ 0 }I } }{ 2\pi r } \left( \pi -\theta +\tan { \theta } \right) $$
  • $$\cfrac { { { \mu }_{ 0 } I} }{ 2\pi r } \left( -\tan { \theta } +\pi -\theta \right) $$
A long wire bent as shown in figure carries current $$I$$. If the radius of the semicircular portion is $$a$$, the magnetic field at center $$C$$ is:

167104.PNG
  • $$\cfrac { { \mu }_{ 0 }I }{ 4a } $$
  • $$\cfrac { { \mu }_{ 0 }I }{ 4\pi a } \sqrt { { \pi }^{ 2 }+4 } $$
  • $$\cfrac { { \mu }_{ 0 }I }{ 4a }+\cfrac { { \mu }_{ 0 }I }{ 4\pi a }$$
  • $$\cfrac { { \mu }_{ 0 }I }{ 4\pi a } \sqrt { { \pi }^{ 2 }-4 } $$
Select the correct statement

167139_4fd145fa686f4ca5ba67285fb0e3b681.png
  • At $$t=\dfrac {T_0}{2}$$, coordinate of charge are $$\left (\dfrac {P_0}{2}, 0, -2R_0\right )$$
  • At $$t=\dfrac {3T_0}{2}$$, coordinate of charge are $$\left (\dfrac {3P_0}{2}, 0, 2R_0\right )$$
  • At $$t=\dfrac {T_0}{2}$$, coordinate of charge are $$\left (P_0, 0, -2R_0\right )$$
  • At $$t=\dfrac {3T_0}{2}$$, coordinate of charge are $$\left (3P_0, 0, 2R_0\right )$$
Figure shows three cases: in all cases the circular part has radius $$r$$ and straight ones are infinitely long. For the same current the ratio of field $$B$$ at center $$P$$ in the three cases $${B}_{1}:{B}_{2}:{B}_{3}$$ is :

166815_2e854f77095442acb80315f91da2e25b.png
  • $$\left( -\cfrac { \pi }{ 2 } \right) :\left( \cfrac { \pi }{ 2 } \right) :\left( \cfrac { 3\pi }{ 4 } -\cfrac { 1 }{ 2 } \right) $$
  • $$\left( -\cfrac { \pi }{ 2 } +1 \right) :\left( \cfrac { \pi }{ 2 } +1 \right) :\left( \cfrac { 3\pi }{ 4 } +\cfrac { 1 }{ 2 } \right) $$
  • $$\left( -\cfrac { \pi }{ 2 } \right) :\left( \cfrac { \pi }{ 2 } \right) :\left( \cfrac { 3\pi }{ 4 } \right) $$
  • $$\left( -\cfrac { \pi }{ 2 } -1 \right) :\left( \cfrac { \pi }{ 2 } -\cfrac { 1 }{ 4 } \right) :\left( \cfrac { 3\pi }{ 4 } +\cfrac { 1 }{ 2 } \right) $$
An otherwise infinite, straight wire has two concentric loops of radii $$a$$ and $$b$$ carrying equal currents in opposite directions as shown in figure. The magnetic field at the common center is zero for:

166818.PNG
  • $$\cfrac { a }{ b } =\cfrac { \pi -1 }{ \pi } $$
  • $$\cfrac { a }{ b } =\cfrac { \pi } { \pi +1 }$$
  • $$\cfrac { a }{ b }=\cfrac { \pi -1 }{ \pi +1 }$$
  • $$\cfrac { a }{ b }=\cfrac { \pi +1 }{ \pi -1 }$$
The magnetic field at the center of the  circular loop as shown in figure, when a single wire is bent to form a circular loop and also extends to form straight sections, is :

166813_df485782256b42b2b4a4dbb1144596fd.png
  • $$\cfrac { { { \mu }_{ 0 } I} }{ 2R } \bigodot $$
  • $$\cfrac { { { \mu }_{ 0 } I} }{ 2R }\left( 1+\cfrac { 1 }{ \pi \sqrt { 2 } } \right) \bigodot $$
  • $$\cfrac { { { \mu }_{ 0 } I} }{ 2R }\left( 1-\cfrac { 1 }{ \pi \sqrt { 2 } } \right) \bigotimes $$
  • $$\cfrac { { { \mu }_{ 0 } I} }{ R }\left( 1-\cfrac { 1 }{ \pi \sqrt { 2 } } \right) \bigotimes $$
Positive point chages $$q=+8.00\mu C$$ and $$q'=+3.00\mu C$$ are moving relative to an observer at point $$P$$, as shown in the figure. the distance $$d$$ is $$0.120$$ $$m$$. When the two charges are at the locations as shown in the figure, what are the magnitude and direction of the net magnetic field they produce at point $$P$$?
(Take $$v=4.50\times { 10 }^{ 6 }{ m{ s }^{ -1 }}$$ and $$v'=9.00\times { 10 }^{ 6 }{ m{ s }^{ -1 }}$$).

167232.png
  • $$4.38\times { 10 }^{ -4 }$$ $$T$$, into the page
  • $$4.38\times { 10 }^{ -4 }$$ $$T$$, out of the page
  • $$2.16\times { 10 }^{ -4 }$$ $$T$$, into the page
  • $$2.9\times { 10 }^{ -9 }$$ $$T$$, out of the page
Figure shows an Amperian path $$ABCDA$$. Part $$ABC$$ is in verical plane $$PSTU$$ while part $$CDA$$ is in horizontal plane $$PQRS$$. Direction of circulation along the path is shown by an arrow near point $$B$$ and $$D$$.
$$\oint { \vec { B } .d\vec { l }  } $$ for this path according to Ampere's law will be :


167202_0acd2e30dfc54ad59a896586006da0a7.png
  • $$\left( { I }_{ 1 }-{ I }_{ 2 }+{ I }_{ 3 } \right) { \mu }_{ 0 }$$
  • $$\left( -{ I }_{ 1 }+{ I }_{ 2 } \right) { \mu }_{ 0 }$$
  • $${ I }_{ 3 }{ \mu }_{ 0 }$$
  • $$\left( { I }_{ 1 }+{ I }_{ 2 } \right) { \mu }_{ 0 }$$
0:0:1


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Practice Class 12 Medical Physics Quiz Questions and Answers