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

4g

Explanation:

Using the right-hand rule, one can determine the direction of magnetic fields on loops $Y$ and $Z$ due to the current in the wire.

The magnetic field on loop $Y$ is in the direction out of the page, and the magnetic field on loop $Z$ is in the direction into the page. Both decrease as the current in the wire decreases, so magnetic flux decreases for both loops $Y$ and $Z$.

According to Lenz' law, a current is induced on the loop $Y$ in an anti-clockwise direction, so it opposes the change in the flux. Likewise, according to Lenz' law, a current is induced on the loop Z in a clockwise direction, so it opposes the change in the flux.

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.

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.

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.

Explanation:

• I. Correct: Since the particle is deflected toward to negative plate it is positively charged.
• II. Correct: In order not to be deflected magnetic force acts on the particle should be upwards. According to the right hand rule the magnetic field should be into the page.
• III. Incorrect: In order not to be deflected magnetic force acts on the particle should be in upwards. According to the right hand rule magnetic field should be into the page.

$Z$ 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:
Conventional current will flow anticlockwise due to the orientation of the cell. To determine the force experienced by a current carrying conductor in a magnetic field, use (any appropriate hand-rule convention or) Flemming's left hand rule which states that if we arrange our thumb, forefinger and middle finger of the left-hand perpendicular to each other, then the thumb points towards the direction of the force experienced by the conductor, the forefinger points towards the direction of the magnetic field and the middle finger points towards the direction of the electric current.

Explanation:
A is incorrect because the equipotentials have equal potential difference but the lines are not equally spaced.

B is incorrect because work =qΔV=(−e)[−30 V−(−50 V)]=−20 eV

C is correct because work =qΔV=(−e)[−30 V−(−10 V)]=20 eV

D is incorrect because work done from R to S is zero (same potential)

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

1/4

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.

Explanation:

Induced emf is proportional with rate of change in magnetic flux. As the distance between the bar magnet and loop magnetic flux changes and the rate of change increases as the relative velocity increases (vL>vK=vM).

Therefore, εL>εK=εM

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.

Explanation:
Choice A is incorrect as it equates the gravitational field strength with the gravitational force and these two are two different physical quantities.

Choice B is the correct answer, since the gravitational field strength equals to force per unit mass; g=F/m.

Choice C is incorrect, because the gravitational field strength is inversely proportional to the square of the distance between the center of mass of the planet and that point; g=GM/r^2.

Choice D is incorrect, because the gravitational field strength depends on the mass of the planet; g=GM/r^2.

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.

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:
The $20nC$ charge produces a leftward electric field in region 1. Therefore, the electric field generated by the $X$ charge should be rightward to cancel the electric field of the 20 nC charge. This is achieved by a negative charge at X. The magnitude of the electric field decreases by distance. All locations in region 1 are closer to point charge $X$ than they are to the $20nC$ charge. Therefore X should be smaller than the charge of $20nC$. The answer is D.

Explanation:
I is correct as the gravitational force is given by F=GMm/r^2

II is incorrect because gravitational field strength is the gravitational acceleration, which is not equivalent to the force.

III is correct because the gravitational force is the only force acting on the satellite and it provides the centripetal force