Explanation:
The direction of the current is perpendicular to the magnetic field in both regions and the magnitude of the field is unchanged. So the magnitude of the force is $F$.

Using Fleming's left hand rule, direction of the force is

Explanation:
The alpha particle has positive charge, therefore its deflection will be in the opposite direction of the electron. (answer A or B)

It has a larger mass, therefore the acceleration will be much smaller, explained by the following calculation:

• qE=ma
• a=qE/m

The alpha particle has twice the charge of the electron, this doubles its acceleration.

It has much more mass than the electron, therefore the denominator in the above ratio increases by a larger factor than the numerator. This means the acceleration of the alpha particle will be much smaller than that of the electron. (its motion will change at a slower rate).

This can also be explained by 2-dimensional motion where there is no force on the horizontal but there is a smaller force on the vertical, meaning it will cover the vertical distance in a longer time, allowing it to travel further along the horizontal, giving the path in B.

• ε=BvL
• power = ε^2/R=[(BvL)^2]/R=B^2v^2L^2/R

The electric field at X due to -Q points toward -Q. The field at X due to +Q points away from +Q, with the same magnitude at X as the field from -Q. Adding the electric field vectors gives a vector pointing to the left.

Explanation:
Summing the four electric field vectors from the individual point charges, the fields from the charge at the top left and the bottom right will cancel out. The fields from the bottom left and top right will add together to give a net field in the direction of the charge in the top right.

$Z$ and $X$ are on the same equipotential line, and $Y$ and $W$ are on the same equipotential line, therefore the magnitudes of the changes in potential will the same.

Moving a particle from $Y$ to $X$ requires −10 eV of work, which means that +10 eV of work is done on the beta-minus particle by the field.

Y to $X$ and $Z$ to $W$ are opposite paths to each other, so the work done by the field on the same particle will be the negative of the work done by the field in the first situation, therefore −10 eV.

Explanation:

When the bar magnet moves towards and away from the ring, the magnetic flux through the ring will change. According to Lenz's law, an emf, and therefore a current is induced to oppose the change in the flux. So as the bar magnet approaches the metal ring, the metal ring moves to the right due to repulsive forces that resist the decrease in distance between the ring and the magnet. Likewise, as the bar magnet moves away from the metal ring, the metal ring moves to the left.

Explanation:
The electric field strength direction is the same as the direction of the electric force.

The electric field will be directed out of the rod in all directions, as it contains uniformly distributed positive charges.

Getting the vector sum for the electric field, due to symmetry the total electric field strength will be directed out of the rod with an angle away from the midpoint of the rod as shown

Also we can think of B as the only correct solution, because it’s a positively charged rod, so the direction must be out of the rod.

1/4

Explanation:
The pattern of equipotential surfaces belongs to two like charges. That indicates that field direction at points A and B should be in the same direction. Additionally, field lines and equipotential surfaces are at right angles to each other. The only answer that satisfies these conditions is A. Therefore, the answer is A.

According to Fleming’s left hand rule, the electrons in the rod experience a force towards Side B at the top of the rod. The flow of electrons (negative charges) towards the top end of the rod will stop when the electrons at that end are numerous enough to push new electrons back by electrostatic repulsion. That is, an electric field is established in the rod where the top is negative and the bottom is positive. The corresponding field is from the bottom to the top (Side D to Side B)

Explanation:
The force acting on the electron is downward at entry and perpendicular to the velocity, and the electron has negative charge. Using conventional methods (right-hand/left-hand) the magnetic field vector points into the plane of the paper

Explanation:
Equipotential lines are perpendicular to field lines, which means the field lines must be directed from A to B or vice versa. That is only possible for 2 charges with opposite sign. In I and II, the forces from the two objects would be in opposite directions, giving a resultant field zero at the midpoint, and equipotential lines of a “peanut” shape.

Explanation:

Because the electric field lines point toward the charge $q$, this charge must be negative, meaning that statement I is false.

The relatively greater number of field lines around charge $q$ means that this charge magnitude is greater than the charge +Q, indicating that statement II is also untrue.

Electric field line density is proportional to electric field strength. Because the field line density near $q$ is greater than that near $+Q$, statement III is correct.

Therefore the correct answer is option C.

Explanation:
According to Newton's third law, the two objects exert equal and opposite forces on each other.

The equation

• Fg=GMm/r

also shows that the forces on the planets are equal in magnitude.

Explanation:
Centripetal acceleration of an orbiting satellite is the gravitational acceleration of the planet, which is given by

g=GM/r^2

Since the gravitational acceleration is inversely proportional to the square of the orbital radius, the correct graphical representation is given in C

Explanation:
At a level closer to the surface of the planet, the centripetal force increases. In a circular orbit, the centripetal force is provided by the gravitational force, which is proportional to the inverse square of the distance from the center of the planet.

The kinetic energy in a closer orbit is larger since the centripetal force is larger. Making the speed larger is necessary to maintain the orbit, which therefore increases the kinetic energy.

4g