Magnetic Effect of Current MCQ Questions & Answers in Electrostatics and Magnetism | Physics
Learn Magnetic Effect of Current MCQ questions & answers in Electrostatics and Magnetism are available for students perparing for IIT-JEE, NEET, Engineering and Medical Enternace exam.
191.
A circular loop of radius $$R,$$ carrying current $$I,$$ lies in $$x - y$$ plane with its centre at origin. The total magnetic flux through $$x - y$$ plane is
A
directly proportional to $$I$$
B
directly proportional to $$R$$
C
inversely proportional to $$R$$
D
zero
Answer :
zero
The magnetic lines of force created due to current will be in such a way that on $$x - y$$ plane these lines will be perpendicular. Further, these lines will be in circular loops. The number of lines moving downwards in $$x - y$$ plane will be same in number to that coming upwards of the $$x - y$$ plane. Therefore, the net flux will be zero. One such magnetic line is shown in the figure.
192.
Two long straight parallel wires, carrying (adjustable) current $${I_1}$$ and $${I_2},$$ are kept at a distance $$d$$ apart, If the force $$'F'$$ between the two wires is taken as ‘positive’ when the wires repel each other and ‘negative’ when the wires attract each other, the graph showing the dependence of $$'F',$$ on the product $${I_1}\,{I_2},$$ would be :
193.
A beam of electrons is moving with constant velocity in a region having simultaneous perpendicular electric and magnetic fields of strength $$20\,V{m^{ - 1}}$$ and $$0.5\,T$$ respectively at right angles to the direction of motion of the electrons. Then the velocity of electrons must be
A
$$8\,m/s$$
B
$$20\,m/s$$
C
$$40\,m/s$$
D
$$\frac{1}{{40}}m/s$$
Answer :
$$40\,m/s$$
The electron moves with constant velocity without deflection. Hence, force due to magnetic field is equal and opposite to force due to electric field.
194.
A conducting circular loop of radius $$r$$ carries a constant current $$i.$$ It is placed in a uniform magnetic field $${{\vec B}_0}$$ such that $${{\vec B}_0}$$ is perpendicular to the plane of the loop. The magnetic force acting on the loop is
A
$$ir\,{B_0}$$
B
$$2\pi \,ir\,{B_0}$$
C
zero
D
$$\pi \,ir\,{B_0}$$
Answer :
zero
The magnetic field is perpendicular to the plane of the paper. Let us consider two diametrically opposite elements. By Fleming's left hand rule, on element $$AB$$ the direction of force will be leftwards and the magnitude will be
$$dF = I\left( {d\ell } \right)B\sin {90^ \circ } = I\left( {d\ell } \right)B$$
On element $$CD,$$ the direction of force will be towards right on the plane of the paper and the magnitude
will be
$$dF = I\left( {d\ell } \right)B.$$
These two forces will cancel out. NOTE : Similarly, all forces acting on the diametrically opposite elements will cancel out in pair. The net force acting on the loop will be zero.