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Moving Charges and Magnetism

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4 months ago

Moving Charges and Magnetism Physics NCERT Exemplar Solutions Class 12th Chapter Four

A current carrying circular loop of radius R is placed in the x-y plane with centre at the origin. Half of the loop with x >0 is now bent so that it now lies in the y-z plane.
(a) The magnitude of magnetic moment now diminishes.
(b) The magnetic moment does not change.
(c) The magnitude of B at (0,0, z), z>>R increases.
(d) The magnitude of B at (0,0,z), z>>R is unchanged.

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alok kumar singh

Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- (a)

Explanation- magnetic moment, M = niA, when we decrease the radius by half its area is decrease by ¼ so the magnetic moment also.so magnetic moment diminishes.

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4 months ago

Moving Charges and Magnetism Physics NCERT Exemplar Solutions Class 12th Chapter Four

Biot-Savart law indicates that the moving electrons (velocity v) produce a magnetic field B such that
(a) B is perpendicular to v.
(b) B is parallel to v.
(c) it obeys inverse cube law.
(d) it is along the line joining the electron and point of observation.

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alok kumar singh

Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer – (a)

Explanation- in biot savarts law magnetic field is perpendicular to v

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4 months ago

Moving Charges and Magnetism Physics NCERT Exemplar Solutions Class 12th Chapter Four

Two charged particles traverse identical helical paths in a completely oppo-site sense in a uniform magnetic field B = B₀k
(a) They have equal z-components of momenta
(b) They must have equal charges
(c) They necessarily represent a particle, anti-particle pair
(d) The charge to mass ratio satisfy

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alok kumar singh

Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer – (d)

Explanation- it is determined by its pitch

Pitch = $\frac{2 π m v c o s θ}{B q}$ = constant

So charge by mass ratio is also constant

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4 months ago

Moving Charges and Magnetism Physics NCERT Exemplar Solutions Class 12th Chapter Four

A long straight wire carrying current of 25 A rests on a table as shown in figure. Another wire PQ of length 1 m, mass 2.5 g carries the same current but in the opposite direction. The wire PQ is free to slide up and down. To what height will PQ rise?

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alok kumar singh

Contributor-Level 10

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

Explanation- magnetic field produced by a long straight current carrying wire is

B= $\frac{μ_{0} I}{2 π r}$ = $\frac{μ_{0} I}{2 π h}$ , for h height distance

Magnetic force on small conductor is F= ilBsin $θ$ = ilB

Also force, F =mg = $\frac{μ_{0} I^{2}}{2 π r}$ l

h= $\frac{4 π * 10^{- 7} * 25 * 25 * 1}{2 π * 2.5 * 10^{- 3} * 9.8}$ = 51 $*$ 10^-4cm

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4 months ago

Moving Charges and Magnetism Physics NCERT Exemplar Solutions Class 12th Chapter Four

A multirange voltmeter can be constructed by using a galvanometer circuit as shown in figure. We want to construct a voltmeter that can measure 2 V, 20 V and 200 V using a galvanometer of resistance 10 ? and that produces maximum deflection for current of 1 mA. Find R₁, R₂and R₃ that have to be used,

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alok kumar singh

Contributor-Level 10

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

Explanation- applying expression in different situation

For i_G (G+R₁) = 2 , for 2V range

i_G (G+R₁+R₂) = 20 , for 20V range

i_G (G+R₁+R₂+R₃) = 2 , for 200V range

on solving we get R₁= 1990ohm, R₂= 18kohm, R₃= 180k ohm

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4 months ago

Moving Charges and Magnetism Physics NCERT Exemplar Solutions Class 12th Chapter Four

Do magnetic forces obey Newton's third law. Verify for two current elements dl₁= dl i located at the origin and dl₂ = dl j located at(0, R, 0). Both carry current I.

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alok kumar singh

Contributor-Level 10

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

Explanation- In Biot-Savart's law, magnetic field B is parallel (II) to id * r and idl have its direction along the direction of flow of current.

Here, for the direction of magnetic field, At dl2, located at (0, R,0) due to wire d1 is given by B I | dl * r or I *j (because point (0, R,0)lies on y-axis), but I * j= k

So, the direction of magnetic field at d2 is along z-direction.

The direction of magnetic force exerted at d2 because of the first wire along the x-axis

F =i (l $*$ B) i.e., F| (i $* k$ ) or along− j direction.

Therefore, force due to dl1 on

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

Explanation- In Biot-Savart's law, magnetic field B is parallel (II) to id * r and idl have its direction along the direction of flow of current.

Here, for the direction of magnetic field, At dl2, located at (0, R,0) due to wire d1 is given by B I | dl * r or I *j (because point (0, R,0)lies on y-axis), but I * j= k

So, the direction of magnetic field at d2 is along z-direction.

The direction of magnetic force exerted at d2 because of the first wire along the x-axis

F =i (l $*$ B) i.e., F| (i $* k$ ) or along− j direction.

Therefore, force due to dl1 on dl2 is non-zero.

Now, for the direction of magnetic field, At d1, located at (0,0,0) due to wire d2 is given by

B I | dl * r or j *−j (because origin lies on y-direction w.r.t. point (0, R,0).), but j *−j =0.

So, the magnetic field at d1 does not exist.

Force due to dl2 on dl1 is zero.

So, magnetic forces do not obey Newton's third law.

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4 months ago

Moving Charges and Magnetism Physics NCERT Exemplar Solutions Class 12th Chapter Four

An electron enters with a velocity v = v₀i into a cubical region (face parallel to coordinate planes) in which there are uniform electric and magnetic fields. The orbit of the electron is found to spiral down inside the cube in a plane parallel to the x-y plane. Suggest a configuration to fields E and B that can lead to it.

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alok kumar singh

Contributor-Level 10

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

Explanation- considering magnetic field along B=Bk and electron enters with a velocity v=vi into a cubical region

The force on electron using Lorentz force = -e (vi $*$ Bk)= evBi

So the force F= -eE₀ k accelerates along z axis which in turn increases the radius of circular path and hence particle transversed on spiral path.

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4 months ago

Moving Charges and Magnetism Physics NCERT Exemplar Solutions Class 12th Chapter Four

A charged particle of charge e and mass m is moving in an electric field E and magnetic field B. Construct dimensionless quantities and quantities of dimension [T]^-1

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alok kumar singh

Contributor-Level 10

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

Explanation- for a charge particle to move in circular path

Mv²/r = qvB

v/r=w= bq/m

w= v/r = $\frac{L T^{- 1}}{L}$ = T-1

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4 months ago

Moving Charges and Magnetism Physics NCERT Exemplar Solutions Class 12th Chapter Four

A current carrying loop consists of 3 identical quarter circles of radius R, lying in the positive quadrants of the x-y, y-z and z-x planes with their centres at the origin, joined together. Find the direction and magnitude of B at the origin.

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alok kumar singh

Contributor-Level 10

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

Explanation- general formula for magnetic field at center is

B= $\frac{μ_{0} I}{4 π r} (θ)$

B1= $\frac{μ_{0} I}{4 π r} (\frac{π}{2}) = \frac{μ_{0} I}{8 r}$ k for xy plane

B2= $\frac{μ_{0} I}{8 r}$ i for zy plane

B3= $\frac{μ_{0} I}{8 r}$ j for xz plane

After joining together B= $\frac{μ_{0} I}{8 π r}$ (i+j+k)

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4 months ago

Moving Charges and Magnetism Physics NCERT Exemplar Solutions Class 12th Chapter Four

Two long wires carrying current I₁, and I₂are arranged as shown in figure. The one carrying current I₁ is along the x-axis. The other carrying current I₂ is along a line parallel to the y-axis given by x = 0 and z = d. Find the force exerted at O₂ because of the wire along the x-axis.

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alok kumar singh

Contributor-Level 10

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

Explanation – in biot savart law, magnetic field B is parallel to idlr and idl have its direction along the direction of flow of current.

Here, for the direction of magnetic field at O₂, due to wire carrying I₁ current is B| parallel idl $*$ r or I $*$ k, but it is -j

So the direction at O₂ is along y-direction.

The direction of magnetic force exerted at O2 because of the wire along x-axis

F= ilB=j (-j)=0

So the force is zero.

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