Moving Charges and Magnetism

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4 months ago

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A
alok kumar singh

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar

Explanation – here, the condition of magnetic resonance is violated.

When the frequency of radio frequency field were doubled, the time period of the radio frequency field were halved. Therefore, the duration in which particle completes half revolution inside the dees, radio frequency completes the cycle.

Hence, particle will accelerate and decelerate alternatively. So the radius of path in the dees will remain same.

New answer posted

4 months ago

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A
alok kumar singh

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar

Explanation- yes, the magnetic force differ from inertial frame to frame. The magnetic force is frame dependent.

The net acceleration which comes into existing out of this is however, frame independent for inertial frames.

New answer posted

4 months ago

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A
alok kumar singh

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar

Explanation- dW=F.dl=0

As dl = vdt

dW= Fvdt

dW= f.v=0

New answer posted

4 months ago

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A
alok kumar singh

Contributor-Level 10

This is a Long Answer Type Questions as classified in NCERT Exemplar

Explanation- (a) Suppose the five wires A, B, C, D and E be perpendicular to the plane of paper at locations as shown in figure.

Thus, magnetic field induction due to five wires will be represented by various sides of a

closed pentagon in one order, lying in the plane of paper. So, its value is zero.

(b) Since, the vector sum of magnetic field produced by each wire at O is equal to 0.

Therefore, magnetic induction produced by one current carrying wire is equal in

magnitude of resultant of four wires and opposite in direction.

Therefore, the field if current in one of the wi

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4 months ago

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alok kumar singh

Contributor-Level 10

This is a Long Answer Type Questions as classified in NCERT Exemplar

Explanation- iG (G) = (I1-IG) (S1+S2+S3) for I1= 10mA

iG (G+S1) = (I2-IG) (S2+S3) for I2= 100mA

iG (G+S1+S2) = (I3-IG) (S3) for I3= 1A

S1= 1W, S2= 0,1W and S3= 0.01W

New answer posted

4 months ago

0 Follower 5 Views

A
alok kumar singh

Contributor-Level 10

This is a Long Answer Type Questions as classified in NCERT Exemplar

Explanation- B (z) points in the same direction on z-axis and hence, J (L) isa monotonically function of L so B.dl = B.dl as cos 0 = 1

(b) B.dl=μoI but when L  

So B  1/r3

(c) the magnetic field due to a circular current carrying loop of radius in the xy plane with centre at origin at any point lying at a distance of from origin.

 

B= μ0IR22 (Z2+R2)3/2

Z=Rtan θ

dz=Rsec2 θdθ

-Bzdz=μ0I2-π/2π/2cosθdθ = μ0I

New question posted

4 months ago

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New answer posted

4 months ago

0 Follower 7 Views

A
alok kumar singh

Contributor-Level 10

This is a Long Answer Type Questions as classified in NCERT Exemplar

Explanation- B (z) points in the same direction on z-axis and hence, J (L) isa monotonically function of L so B.dl = B.dl as cos 0 = 1

(b) B.dl=μoI but when L  

So B  1/r3

(c) the magnetic field due to a circular current carrying loop of radius in the xy plane with centre at origin at any point lying at a distance of from origin.

 

B= μ0IR22 (Z2+R2)3/2

Z=Rtan θ

dz=Rsec2 θdθ

-Bzdz=μ0I2-π/2π/2cosθdθ = μ0I

New answer posted

4 months ago

0 Follower 1 View

A
alok kumar singh

Contributor-Level 10

This is a Long Answer Type Questions as classified in NCERT Exemplar

Explanation- as we know magnetic moment = nIA

For equilateral triangle M= nIA= 4I ( 34a2 )

M= Ia23

For square, n=3 so total length of wire is 12a

M= nIA= 3I (a2) = 3Ia2

For regular hexagon of side a, n=2 so total length = 12a

M= nIA=2 ( 634a2 )= 3 3 a2I

New answer posted

4 months ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

This is a Long Answer Type Questions as classified in NCERT Exemplar

Explanation- Since, B is along the x-axis, for a circular orbit the momenta of the two particles are in the y-z plane. Let P1 and P2 be the momentum of the electron and positron, respectively. Both traverse a circle of radius R of opposite sense. Let P1 make an angle? with the y-axis P2 must make the same angle.

The centres of the respective circles must be perpendicular to the momenta and at a distance R. Let the centre of the electron be at Ce and of the positron at Cp . The coordinates of Ce is

Ce= (0, -Rsin θ, Rcosθ )

Cp= (0, -Rsin θ, -32R-Rcosθ )

The circles of the two

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