Magnetic Effect of Current — Page 7 | NEET Physics

Q61

Two circular coils 1 and 2 are made from the same wire but the radius of the 1 st coil is twice that of the 2 nd coil. What is the ratio of potential difference in volts should be applied across them so that the magnetic field at their centres is the same?

A 2

B 3

C 4

D 6

Correct Answer

Option C

Solution

Let r 1 and r 2 are the radius of coil 1 and 2. If B 1 and B 2 are magnetic induction at their centre, then

{B_1} = {{{\mu _0}{I_1}} \over {2{r_1}}}

and

{B_2} = {{{\mu _0}{I_2}} \over {2{r_2}}}

Since B 1 = B 2 ; and r 1 = 2r 2 therefore I 1 = 2I 2 .

Again if R 1 and R 2 are resistance of the coil 1 and 2 then R 1 = 2R 2 (as R $\propto$ length = 2 $\pi$ r) and if V 1 and V 2 are the potential difference across them respectively, then

{{{V_1}} \over {{V_2}}} = {{{I_1}{R_1}} \over {{I_2}{R_2}}} = {{\left( {2{I_2}} \right)\left( {2{R_2}} \right)} \over {{I_2}{R_2}}} = 4

Q62

When a charged particle moving with velocity

\overrightarrow v

is subjected to a magnetic field of induction

\overrightarrow B

, the force on it is non-zero. This implies that

A angle between is either zero or 180 o

B angle between is necessarily 90 o

C angle between can have any value other than 90 o

D angle between can have any value other than zero and 180 o .

Correct Answer

Option D

Solution

Force on a particle moving with velocity

\overrightarrow v

in a magnetic field

\overrightarrow B

is

\overrightarrow F = q\left( {\overrightarrow v \times \overrightarrow B } \right)

$\Rightarrow$ F = qvBsin $\theta$ (i) When $\theta$ = 0°, F = qvBsin0° = 0 (ii) When $\theta$ = 90°, F = qvBsin90° = qvB (iii) When $\theta$ = 180°, F = qvBsin180° = 0 If angle between v and B is either zero or 180º, then value of F will be zero.

Q63

A very long straight wire carries a current

I

. At the instant when a charge +Q at point P has velocity

\overrightarrow v

, as shown, the force on the charge is

A along Oy

B opposite to Oy

C along Ox

D opposite to Ox

Correct Answer

Option A

Solution

The direction of

\overrightarrow B

is due to i is acting inwards i.e. into the paper along

\left( { - \widehat k} \right)

. $\therefore$ The magnetic force

\overrightarrow F = Q\left( {\overrightarrow v \times \overrightarrow B } \right) = Q\left( {\overrightarrow v \widehat i} \right) \times B\left( { - \widehat k} \right) = QvB\widehat j

So,

\overrightarrow F

is along Oy direction.

Q64

An electron moves in a circular orbit with a uniform speed v. It producess a magnetic field B at the centre of the circle. The radius of the circle is proportional to

A

\sqrt {B/v}

B

B/v

C

\sqrt {v/B}

D

v/B

Correct Answer

Option C

Solution

The magnetic field produce by moving electron in circular path B =

{{{\mu _0}i} \over {2r}}

and i =

{q \over {2\pi r}} \times v

$\therefore$ B =

{{{\mu _0}qv} \over {4\pi {r^2}}}

$\Rightarrow$ r $\propto$

\sqrt {{v \over B}}

Q65

To convert a galvanometer into a voltmeter one should connect a

A high resistance in series with galvanometer

B low resistance in series with galvanometer

C high resistance in parallel wilh galvanometer

D low resistance in parallel with galvanometer.

Correct Answer

Option A

Solution

For converting galvanometer to voltmeter, a high resistance should be connected in series with galvanometer.

Q66

A galvanometer of 50 ohm resistance has 25 divisions. A current of 4

\times

10

-

4 ampere gives a deflection of one division. To convert this galvanometer into a voltmeter having a range of 25 volts, it should be connected with a resistance of

A

2500\,\Omega

as a shunt

B 2450

\Omega

as a shunt

C 2550

\Omega

in series

D 2450

\Omega

in series

Correct Answer

Option D

Solution

The total current shown by the galvanometer, I g = 25 $\times$ 4 $\times$ 10 -4 A = 10 -2 A The value of resistance connected in series to convert galvanometer into voltmeter of 25 V is V = I g (R e + R g ) $\Rightarrow$ R e =

{V \over {{I_g}}} - {R_g}

= 2450

\Omega

Q67

A charged particle moves through a magnetic field in a firection perpendicular to it. Then the

A speed of the particle remains unchanged

B direction of the particle remains unchanged

C acceleration remains unchanged

D velocity remains unchanged

Correct Answer

Option A

Solution

If a moving charged particle is subjected to a perpendicular uniform magnetic field, then according to F = qvB sin $\theta$ , it will experience a maximum force which will provide the centripetal force to particle and it will describe a circular path with uniform speed.

Q68

A long solenoid carrying a current producess a magnetic field B along its axis. If the current is doubled and the number of turns per cm is halved, the new value of the magnetic field is

A B/2

B B

C 2B

D 4B

Correct Answer

Option B

Solution

Magnetic field induction at point inside the solenoid of length l, having n turns per unit length carrying current i is given by B = $\mu$ 0 ni If i $\to$ doubled, n $\to$ halved then B $\to$ remains same.

Q69

The magnetic field of given length of wire for single turn coil at its centre is B then its value for two turns coil for the same wire is

A B/4

B B/2

C 4B

D 2B

Correct Answer

Option C

Solution

B 1 = B =

{{{\mu _0}I} \over {2R}}

Further, B 2 =

{{{\mu _0}\left( {2I} \right)} \over {2r}}

Now 2 × 2 $\pi$ r = 2 $\pi$ R or r = R/2 Hence B 2 =

4 \times {{{\mu _0}I} \over {2R}} = 4B

Q70

A charge q moves in a region where electric field and magnetic field both exist, then force on it is

A

q\left( {\overrightarrow v \times \overrightarrow B } \right)

B

q\overrightarrow E + q\left( {\overrightarrow v \times \overrightarrow B } \right)

C

q\overrightarrow E + \overrightarrow q \left( {\overrightarrow B \times \overrightarrow v } \right)

D

q\overrightarrow B + \overrightarrow q \left( {\overrightarrow E \times \overrightarrow v } \right)

Correct Answer

Option B

Solution

Lorentz forece

\overrightarrow {{F_L}} = \overrightarrow {{F_e}} + \overrightarrow {{F_m}}

=

q\overrightarrow E + q\left( {\overrightarrow v \times \overrightarrow B } \right)