Physics Ncert Solutions Class 12th

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

B= $\frac{{\mu }_{0}I}{2\pi r}$ = $\frac{{\mu }_{0}I}{2\pi h}$ , for h height distance

Magnetic force on small conductor is F= ilBsin $\theta$ = ilB

Also force, F =mg = $\frac{{\mu }_0{I}^2}{2\pi r}$ l

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

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

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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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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Explanation- general formula for magnetic field at center is

B= $\frac{{\mu }_{0}I}{4\pi r}\left(\theta \right)$

B1= $\frac{{\mu }_{0}I}{4\pi r}\left(\frac{\pi }{2}\right)=\frac{{\mu }_{0}I}{8r}$ k for xy plane

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

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

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

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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.

This is a short answer type question as classified in NCERT Exemplar

No, the potential function have no free space because when there is a free space there will be a leakage of potential at that point.

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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.

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It is only possible when their sizes are different . when their sizes are different there capacity would change. So there is potential difference btween them.