A circular coil of 14 turns of radius 5 cm is in a uniform magnetic field of 4000 G in the positive x direction. Find the flux through the coil for the following orientations of the unit vector n that is perpendicular to the plane of the coil. (a) n = i Wb [2 points] 0 attempt(s) made (maximum allowed for credit = 5) [after that, multiply credit by 0.5 up to 10 attempts] (b) n = j Wb [2 points] 0 attempt(s) made (maximum allowed for credit = 5) [after that, multiply credit by 0.5 up to 10 attempts] (c) n = (i + j)/square root of 2 Wb [2 points] 0 attempt(s) made (maximum allowed for credit = 5) [after that, multiply credit by 0.5 up to 10 attempts] (d) n = k Wb [2 points] 0 attempt(s) made (maximum allowed for credit = 5) [after that, multiply credit by 0.5 up to 10 attempts] (e) n = 0.7 i + 0.714142842854285 j Wb

The question is not clear.

Complete question is;

A circular coil of 14 turns of radius 5 cm is in a uniform magnetic field of 4000 G in the positive x direction. Find the flux through the coil for the following orientations of the unit vector n that is perpendicular to the plane of the coil.

(a) n = i Wb

(b) n = j Wb

(c) n = (i + j)/√2 Wb

(d) n = k Wb

(e) n = 0.7 i + 0.714142842854285 j Wb

A) Electric flux = 0.044Wb

B) Electric flux = 0 Wb

C) Electric flux = 0.031 Wb

D) Electric flux = 0 Wb

E) Electric flux = 0.0308 Wb

Explanation:

equation for magnetic flux is given as;

Φ = NBAcosθ

where;

Φ is the magnetic flux

θ is the angle between n and B

N is the number of turns of wire.

A is the area.

Now,

N = 14 turns

Radius(r) = 5cm = 0.05 m

Area(A) = πr² = π x 0.05² =0.007854 m²

Magnetic field (B) = 4000G = 0.4 T

A) we are given the vector n = i Wb

Thus following the pattern of i + j used, where the number attached to j is y-value while that attached to i is x-value. So, y/x = tanθ

So, θ = tan^-1(y/x)

In this case, there is no j, so y = 0 and x = 1. so,

tanθ = 0/1 = 0 ;θ = tan^-1(0) =0

Thus,

Φ = NBAcosθ = 14 x 0.4 x 0.007854 x cos 0 = 0.044Wb

B)we are given the vector n = i Wb

Following the pattern in the above,

tan(θ) = 1/0

This doesn't exist and thus, magnetic flux = 0Wb

C) we are given the vector;

n = (i + j)/√2 Wb

So, Following the pattern in the first example,

tan(θ) = (1/√2)/(1/√2) = 1

θ = tan^-1(1) =45°

Thus,

Φ = NBAcosθ = 14 x 0.4 x 0.007854 x cos 45 = 0.031 Wb

D) n = k Wb

K is a neutral vector and as such Φ = 0

E) we are given;

n = 0.7 i + 0.714142842854285 j Wb

So, Following the pattern in the first example,

tan(θ) = (0.714142842854285)/(0.7) = 1.0202

θ = tan^-1(1.0202) =45.57°

Thus,

Φ = NBAcosθ = 14 x 0.4 x 0.007854 x cos 45.57 = 0.0308 Wb

answer: 2×10^-2kg

explanation: density= mass per volume

vol=mass/density

converting to s.i units we have;

3÷1000kg = 0.003

1.5÷100m = 0.15

0.003/0.15 = 0.02kg

weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards