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

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​:
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​:
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​:
The strong nuclear force is the strongest of the four fundamental forces, then comes electromagnetic, then weak nuclear force and finally gravitation

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​:

• 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.

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​:
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​:
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.

$Z$ 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​:
The negative ion is attracted by the positive charge, therefore its energy at the inner equipotential line is less than the outer one.

Moving from 1 to 2, it moves to a lower energy, therefore work is negative

Moving from 2 to 3, it moves along the same equipotential line, therefore no work done

Moving from 3 to 4, it moves to a higher energy, therefore positive work is done.

Explanation​:
The electric field vector inside a hollow conductor is zero. Due to the lack of field inside the sphere, the electric potential is constant and equal to the surface value . Outward from the surface of the sphere, the electric field varies with the inverse​ square of the distance from the center while electric potential decreases inversely with the distance from the center.

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​:
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

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.

4g

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.