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$A circular coil of radius R is carrying a certain current . At a point of its axis at distance x from center the value of magnetic field is found to be \frac{1}{2 \sqrt{2}} times the magnetic field at its center . The value of x must be :$

$\frac{R}{2}$

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Explanation

$(1) \frac{μ_{0}}{2} \frac{{NIR}^{2}}{{(R^{2} + x^{2})}^{3 / 2}} = \frac{1}{2 \sqrt{2}} \frac{μ_{0} NI}{2 R} \Rightarrow x = R$

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When positively charged particle falling vertically downward, then the particle deflected due to earth magnetic field towards

east

west

north

south

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Explanation

Using Fleming left-hand rule

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The AC voltage across a resistance can be measured using a :

Potentiometer

Hotwire voltmeter

Moving coil galvanometer

Moving magnet galvanometer

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Explanation

2. Hotwire voltmeter

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The magnetic field at center due to the orbital motion of the electron in the hydrogen atom is :

12.5 tesla

5 tesla

24 tesla

1 tesla

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Explanation

$(1) Magnetic field for hydrogen like atom is B = \frac{12.5 z^{2}}{n^{2}} tesla \Rightarrow B = \frac{12.5 \times 1^{2}}{1^{2}} = 12.5 t e s l a$

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A galvanometer of resistance 240 $Ω$ allows only 4% of the main current after connecting a shunt resistance. The value of shunt resistance is :

10 $Ω$

20 $Ω$

8 $Ω$

5 $Ω$

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Explanation

$(1) I_{g} = \frac{4}{100} I (given) S = \frac{I_{g} G}{I - I_{g}} = 10 Ω$

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A circular coil of wire of radius 'r' has 'n' turns and carries a current 'I'. The magnetic induction (B) at a point on the axis of the coil at a distance $\sqrt{3} r$ from its center is :

$\frac{μ_{0} In}{4 r}$

$\frac{μ_{0} In}{8 r}$

$\frac{μ_{0} In}{16 r}$

$\frac{μ_{0} In}{32 r}$

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Explanation

$B = \frac{μ_{0}}{4 π} \frac{2 {πr}^{2} \times I \times n}{{(x^{2} + r^{2})}^{3 / 2}} B = \frac{μ_{0}}{2} \frac{r^{2} \times I}{{(r^{2} + 3 r^{2})}^{3 / 2}} \times n = \frac{μ_{0}}{16} \frac{In}{r}$

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A long solenoid has 800 turns per meter length of the solenoid. A current of 1.6 A flows through it. The magnetic induction at the end of the solenoid on its axis is:

$16 \times 10^{- 4} T$

$8 \times 10^{- 4} T$

$32 \times 10^{- 4} T$

$4 \times 10^{- 4} T$

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Explanation

$B_{end} = \frac{μ_{0} ni}{2} = \frac{4 π \times 10^{- 7} \times 800 \times 1.6}{2} = 8 \times 10^{- 4} T$

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The unit of reduction factor of the tangent galvanometer is

Ampere

Gauss

Radian

None of these

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Explanation

In tangent galvanometer (used to measure small current),

i = k $\tan θ$

where k is the reduction factor

So, by dimensional analysis,

[k] = Current

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The resistance of 1 A ammeter is 0.018 Ω. To convert it into 10 A ammeter, the shunt resistance required will be :

0.18 Ω

0.0018 Ω

0.002 Ω

0.12 Ω

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Explanation

$S = \frac{i_{g} G}{(i - i_{g})} = \frac{1 \times 0 .018}{10 - 1} = \frac{0 .018}{9} = 0 .002 Ω$

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In order to pass 10% of the main current through a moving coil galvanometer of 99 ohms, the resistance of the required shunt is :

9.9 Ω

10 Ω

11 Ω

9 Ω

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Explanation

Shunt resistances $S = \frac{i_{g} G}{(i - i_{g})} = \frac{10 \times 99}{(100 - 10)} = 11 Ω$

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