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UNO [17]
3 years ago
8

A circular loop of wire is held in a uniform magnetic field, with the plane of the loop perpendicular to the field lines. Which

of the following will not cause a current to be induced in the loop?
(a) crushing the loop
(b) rotating the loop about an axis perpendicular to the field lines
(c) keeping the orientation of the loop fixed and moving it along the field lines
(d) pulling the loop out of the field
Physics
1 answer:
frutty [35]3 years ago
4 0

Answer:

(c) keeping the orientation of the loop fixed and moving it along the field lines

Explanation:

Ampere's Law current is calculated by evaluating;

∫ B. dl

Here B= magnetic field and dl- length segment in which current is to flow.

Clearly as magnate field lines are perpendicular to the plane of loop having some length ==> dot product will be zero for 90 ° angle and hence current will be zero in the loop i-e we will keep the position of loop and move it along field lines !

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A 100g block lies on an inclined plane that makes an angle of 15 degrees with the horizontal. The coefficient of kinetic frictio
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Mass that one should put in the container so that the 100 g block slides down the inclined plane at constant speed = 34.16 g

Explanation:

The vertical forces (with respect to the inclined plane) acting on the 100 g block include the component of the weight of the block in the direction vertical to the inclined plane and the normal reaction of the plane on the block.

And sum of upward forces = sum of downward forces.

N = mg cos θ

m = 100 g = 0.10 kg

g = acceleration due to gravity = 9.8 m/s²

θ = 15°

N = (0.1×9.8×cos 15°) = 0.946582 N

The horizontal forces (With respect to the inclined plane) include the frictional force (acting upwards for the inclined plane, opposite to the intended direction of motion), the Tension in the rope (acting downwards, away from the 100 g block) and the horizontal component (with respect to the inclined plane) of the weight of the block, F, (also acting downards).

For the body to slide down the inclined plane at constant speed, the downward sloping forces must balance the frictional force, that is, there will be no acceleration.

Frictional force = Tension + F

Frictional force = μN

where μ = coefficient of kinetic friction = 0.60

N = normal reaction = 0.9466 N

Frictional force = Fr = (0.60 × 0.9466) = 0.56796 N = 0.568 N

The horizontal component (with respect to the inclined plane) of the weight of the block (also acting downards) = mg sin θ

F = (0.10 × 9.8 × sin 15°) = 0.253624 N

Tension in the rope = T = ?

Fr = F + T

T = Fr - F = 0.568 - 0.253624 = 0.314376 N = 0.3144 N

But the balance on the rope now has the total weight on the container (weight of container + weight on the container) to be equal to 2T.

2T = mg

2 × 0.3144 = 9.8m

m = 0.06416 kg = 64.16 g.

Mass of the container = 30 g

So, mass that one should put in the container so that the 100 g block slides down the inclined plane at constant speed = 64.16 - 30 = 34.16 g

Hope this Helps!!!

8 0
3 years ago
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