1
MCQ (Single Correct Answer)

JEE Main 2016 (Online) 10th April Morning Slot

Consider a thin metallic sheet perpendicular to the plane of the paper moving with speed ‘v’ in a uniform magnetic field B going into the plane of the paper (See figure). If charge densities $$\sigma $$1 and $$\sigma $$2 are induced on the left and right surfaces, respectively, of the sheet then (ignore fringe effects) :

A
$$\sigma $$1 = $$ \in $$0 $$\upsilon $$ B, $$\sigma $$2 = $$-$$ $$ \in $$0 $$\upsilon $$ B
B
$$\sigma $$1 = $${{{ \in _0}\upsilon \,B} \over 2},$$ $$\sigma $$2 = $${{ - { \in _0}\,\upsilon B} \over 2}$$
C
$$\sigma $$1 = $$\sigma $$2 = $${ \in _0}\,\upsilon B$$
D
$$\sigma $$1 = $${{ - { \in _0}\upsilon B} \over 2},$$ $$\sigma $$2 = $${{ { \in _0}\upsilon B} \over 2},$$

Explanation

Magnetic force on electron

$$\overrightarrow F $$ = $$-$$ e $$\left( {\overrightarrow V \times \overrightarrow B } \right)$$

F = eVB   [As $${\overrightarrow V }$$ and $${\overrightarrow B }$$ are perpendicular]

Also,   F = e E

and  E = $${\sigma \over {{\varepsilon _0}}}$$

$$ \therefore $$   eVB = e $$ \times $$ $${\sigma \over {{\varepsilon _0}}}$$

$$ \Rightarrow $$   $$\sigma $$ = $${{\varepsilon _0}}$$VB = $$\sigma $$1

as   $$\sigma $$1 = $$-$$ $$\sigma $$2

$$ \therefore $$   $$\sigma $$2 = $$-$$ $${{\varepsilon _0}}$$VB
2
MCQ (Single Correct Answer)

JEE Main 2016 (Online) 10th April Morning Slot

A galvanometer has a 50 division scale. Battery has no internal resistance. It is found that there is deflection of 40 divisions when R = 2400 $$\Omega $$. Deflection becomes 20 divisions when resistance taken from resistance box is 4900 $$\Omega $$. Then we can conclude :

A
Resistance of galvanometer is 200 $$\Omega $$
B
Full scale deflection current is 2 mA.
C
Current sensitivity of galvanometer is 20 $$\mu $$A/division.
D
Resistance required on R.B. for a deflection of 10 divisions is 9800 $$\Omega $$.

Explanation

Let full scale deflection of current = I

In case 1, when R = 2400 $$\Omega $$ and deflection of 40 divisions present.

$$ \therefore $$   $${{40} \over {50}}{\rm I} = {V \over {G + R}}$$

$$ \Rightarrow $$   $${4 \over 5}{\rm I}$$ = $${2 \over {G + 2400}}$$          . . .(1)

Incase 2, when R = 4900 $$\Omega $$ and deflection of 20 divisions present

$$ \therefore $$   $${{20} \over {50}}{\rm I} = {V \over {G + R}}$$

$$ \Rightarrow $$   $${2 \over 5}{\rm I}$$ = $${2 \over {G + 4900}}$$          . . .(2)

From (1) and (2) we get,

$${4 \over 2} = {{G + 4900} \over {G + 2400}}$$

$$ \Rightarrow $$   2G + 4800 $$=$$ G + 4900

$$ \Rightarrow $$   G $$=$$ 100 $$\Omega $$

Putting value of G in equation (1), we get,

$${4 \over 5}{\rm I} = {2 \over {100 + 2400}}$$

$$ \Rightarrow $$   $${\rm I} = 1$$ mA

Current sensitivity $$=$$ $${{\rm I} \over {number\,\,of\,\,divisions}}$$

$$=$$ $${1 \over {50}}$$

$$=$$ 0.02 mA / division

$$=$$ 20 $$\mu $$A / division

Resistance required for deflection of 10 divisions

$${{10} \over {50}}{\rm I} = {V \over {G + R}}$$

$$ \Rightarrow $$   $${1 \over 5} \times 1 \times {10^{ - 3}} = {2 \over {100 + R}}$$

$$ \Rightarrow $$   R $$=$$ 9900 $$\Omega $$
3
MCQ (Single Correct Answer)

JEE Main 2017 (Offline)

When a current of 5 mA is passed through a galvanometer having a coil of resistance 15$$\Omega $$, it shows full scale deflection. The value of the resistance to be put in series with the galvanometer to convert it into a voltmeter of range 0 – 10V is:
A
4.005 × 103 $$\Omega $$
B
1.985 × 103 $$\Omega $$
C
2.535 × 103 $$\Omega $$
D
2.045 × 103 $$\Omega $$

Explanation

Given : Current through the galvanometer,

ig = 5 × 10–3 A

Galvanometer resistance, G = 15 $$\Omega $$

Let resistance R to be put in series with the galvanometer to convert it into a voltmeter.

V = ig (R + G)

10 = 5 × 10–3 (R + 15)

$$ \therefore $$ R = 2000 – 15 = 1985

= 1.985 × 103 $$\Omega $$
4
MCQ (Single Correct Answer)

JEE Main 2017 (Offline)

A magnetic needle of magnetic moment 6.7 $$\times$$ 10-2 A m2 and moment of inertia 7.5 $$\times$$ 10-6 kg m2 is performing simple harmonic oscillations in a magnetic field of 0.01 T. Time taken for 10 complete oscillations is:
A
8.76 s
B
6.65 s
C
8.89 s
D
6.98 s

Explanation

Given : Magnetic moment, M = 6.7 × 10–2 Am2

Magnetic field, B = 0.01 T

Moment of inertia, I = 7.5 × 10–6 Kgm2

Using, T = $$2\pi \sqrt {{I \over {MB}}} $$

= $$2\pi \sqrt {{{7.5 \times {{10}^{ - 6}}} \over {6.7 \times {{10}^{ - 2}} \times 0.01}}} $$

= $${{2\pi } \over {10}} \times 1.06$$ s

Time taken for 10 complete oscillations

t = 10T = 2$$\pi $$ × 1.06

= 6.6568 $$ \simeq $$ 6.65 s

Questions Asked from Magnetics

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