Unit 12: Magnetism and Electromagnetic Induction
Weightage : 15–18%
A beam of electrons travels between two parallel wire coils, as shown in the figures above. When the coils carry no current, the electron beam is undeflected and hits the center of the screen, as indicated by the dashed line. When a constant current \(I\) in the indicated direction is created in the coils, the electron beam is deflected toward which edge of the screen?
Correct Answer: \( \boxed{\mathrm{B}} \)
The two parallel wire coils produce a nearly uniform magnetic field between them. Using the right-hand rule for the current in the coils, the magnetic field between the coils is directed toward the left in the top view.
The magnetic force on a moving charge is
\( \vec{F}=q\vec{v}\times\vec{B} \)
Because the particles are electrons, their charge is negative. Therefore, the force is opposite to the direction given by the right-hand rule for a positive charge.
Applying the right-hand rule to the electron beam and accounting for the negative charge shows that the magnetic force is directed downward.
Thus, the electron beam is deflected toward the bottom of the screen.
Therefore, the correct answer is (B).
A circuit consists of a battery and some wire with significant resistance. The wire is bent so that the circuit is in the shape of a square, and the circuit is aligned in the plane of the page, as shown above. Which of the following best describes the direction of the magnetic field near point \(P\)?
Correct Answer: \( \boxed{\mathrm{C}} \)
Conventional current flows clockwise around the square loop, as determined by the battery polarity.
Using the right-hand rule, curl the fingers of your right hand in the direction of the current. Your thumb points in the direction of the magnetic field through the center of the loop.
For a clockwise current, the magnetic field at point \(P\) is directed into the page. The contributions from all four sides of the square reinforce each other at the center.
Key Formula: \(B=\dfrac{\mu_0 I}{2\pi r}\) (magnetic field due to a long straight current-carrying wire). The net field at \(P\) is found by applying the right-hand rule to each segment of the loop.
Therefore, the correct answer is (C).
Two long, straight, parallel wires in the plane of the page carry equal currents \(I\) in the same direction, as shown above. Which of the following correctly describes the forces acting on the wires and the resultant magnetic field at points along the dotted line midway between the wires?
Correct Answer: \( \boxed{\mathrm{B}} \)
Two parallel current-carrying wires with currents in the same direction exert an attractive magnetic force on each other.
Using the right-hand rule, the magnetic field produced by the upper wire at the midpoint is directed into the page, while the field produced by the lower wire is directed out of the page.
Since the currents are equal and the midpoint is equidistant from both wires, the magnetic fields have equal magnitudes:
\( B=\dfrac{\mu_{0}I}{2\pi r} \)
Therefore, the two fields cancel each other, giving a resultant magnetic field of zero along the dotted line midway between the wires.
Thus, the wires attract each other, and the net magnetic field at the midpoint is zero.
Therefore, the correct answer is (B).
