Two equal electric currents are flowing perpendicular to each other as shown in the figure. $$AB$$ and $$CD$$ are perpendicular to each other and symmetrically placed w.r.t the currents, where do we expect the resultant magnetic field to be zero?
A.
On $$AB$$
B.
On $$CD$$
C.
On both $$AB$$ and $$CD$$
D.
On both $$OD$$ and $$BO$$
Answer :
On $$AB$$
Solution :
Applying right hand grip rule and considering $$AB,$$ the direction of magnetic field due to one current is upwards and that due to other is downwards. Both the magnetic fields cancel out each other and the resultant magnetic field is zero.
Considering $$CD$$ and applying right hand grip rule for the two currents, the direction of magnetic field is in the same direction in both the cases giving non-zero resultant.
Releted MCQ Question on Electrostatics and Magnetism >> Magnetic Effect of Current
Releted Question 1
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 battery is connected between two points $$A$$ and $$B$$ on the circumference of a uniform conducting ring of radius $$r$$ and resistance $$R.$$ One of the arcs $$AB$$ of the ring subtends an angle $$\theta $$ at the centre. The value of the magnetic induction at the centre due to the current in the ring is
A.
proportional to $$2\left( {{{180}^ \circ } - \theta } \right)$$
A proton, a deuteron and an $$\alpha - $$ particle having the same kinetic energy are moving in circular trajectories in a constant magnetic field. If $${r_p},{r_d},$$ and $${r_\alpha }$$ denote respectively the radii of the trajectories of these particles, then
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