Physics NCERT Exemplar Solutions Class 12th Chapter Four

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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.

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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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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.

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Explanation- dW=F.dl=0

As dl = vdt

dW= Fvdt

dW= f.v=0

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Explanation- i_G (G) = (I₁-I_G) (S₁+S₂+S₃) for I₁= 10mA

i_G (G+S₁) = (I₂-I_G) (S₂+S₃) for I₂= 100mA

i_G (G+S₁+S₂) = (I₃-I_G) (S₃) for I₃= 1A

S₁= 1W, S₂= 0,1W and S₃= 0.01W

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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) $\int B.dl={\mu }_{o}I$ but when L $\to$ $\infty$

So B $\propto$ 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= $\frac{{\mu }_{0}I{R}^2}{2{(Z^2+R^2)}^{3/2}}$

Z=Rtan $\theta$

dz=Rsec² $\theta d\theta$

${\int }_{-\infty }^{\infty }{B}_{z}dz=\frac{{\mu }_{0}I}{2}{\int }_{-\pi /2}^{\pi /2}cos\theta d\theta$ = ${\mu }_{0}I$

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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) $\int B.dl={\mu }_{o}I$ but when L $\to$ $\infty$

So B $\propto$ 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= $\frac{{\mu }_{0}I{R}^2}{2{(Z^2+R^2)}^{3/2}}$

Z=Rtan $\theta$

dz=Rsec² $\theta d\theta$

${\int }_{-\infty }^{\infty }{B}_{z}dz=\frac{{\mu }_{0}I}{2}{\int }_{-\pi /2}^{\pi /2}cos\theta d\theta$ = ${\mu }_{0}I$

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 ( $\frac{\sqrt[]{3}}{4}{a}^2$ )

M= Ia²$\sqrt[]{3}$

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 ( $\frac{6\sqrt[]{3}}{4}{a}^2$ )= 3 $\sqrt[]{3}$ a²I