A magnetic field of 4 × 10 - 3 k ^ tesla exerts a force of 4 i ^ + 3 j ^ × 10 - 10 newton on a particle having a charge of 10 - 9 coulumb in X - Y plane Find the velocity of the particle.

Given Magnetic field B = 4 × 10 - 3 k ^ T Force F = 4 i ^ + 3 j ^ × 10 - 10 N Charge on the particle q = 10 - 9 C Considering the motion of the charged particle in X - direction The force on a moving charged particle is given by F = qVB Where q = charge, V = velocity of the particle and B = magnetic field. 4 i ^ + 3 j ^ × 10 - 10 = 10 - 9 V x i ^ + v y j ^ + v z k ^ 4 × 10 - 3 k ^ F x = qV y B or V y = F x qB = 4 × 10 - 10 1 × 10 - 3 × 4 × 110 - 3 = 100 m / s In Y - direction F y = qV x B or V x = F y qB = 3 × 10 - 10 1 × 10 - 3 × 4 × 110 - 3 = 75 m / s Therefore, the velocity V = V x i ^ + V y j ^ = - 75 i ^ + 100 j ^ ms - 1

A magnetic field of 4 × 10 - 3 k ^ tesla exerts a force of 4 i ^ + 3 j ^ × 10 - 10 newton on a particle having a charge of 10 - 9 coulumb in X - Y plane Find the velocity of the particle.

Given Magnetic field B = 4 × 10 - 3 k ^ T Force F = 4 i ^ + 3 j ^ × 10 - 10 N Charge on the particle q = 10 - 9 C Considering the motion of the charged particle in X - direction The force on a moving charged particle is given by F = qVB Where q = charge, V = velocity of the particle and B = magnetic field. 4 i ^ + 3 j ^ × 10 - 10 = 10 - 9 V x i ^ + v y j ^ + v z k ^ 4 × 10 - 3 k ^ F x = qV y B or V y = F x qB = 4 × 10 - 10 1 × 10 - 3 × 4 × 110 - 3 = 100 m / s In Y - direction F y = qV x B or V x = F y qB = 3 × 10 - 10 1 × 10 - 3 × 4 × 110 - 3 = 75 m / s Therefore, the velocity V = V x i ^ + V y j ^ = - 75 i ^ + 100 j ^ ms - 1

In the Bohr model of the hydrogen atom, the electron circulates around the nucleus in the path of radius 5 . 1 × 10 11 m at a frequency of 6 . 8 × 10 15 Hz . Calculate the magnetic induction B at the centre of the orbit. What is the equivalent dipole moment?

Given the radius of the electron orbit r = 5 . 1 × 10 - 11 m Frequency of the electron f = 6 . 8 × 10 15 Hz . The equivalent current is I = e × f = 1 . 6 × 10 - 19 = 1 . 09 × 10 - 3 A The magnetic field due to this current at the centre of the orbit B = μ 0 × I 2 r = 4 × 10 - 7 × 1 . 09 × 10 - 3 2 × 5 . 1 × 10 - 11 = 13 . 4 web / m 2 The magnetic dipole moment = current × area of the orbit = 1 . 09 × 10 - 3 π 5 . 1 × 10 - 11 2 = 8 . 82 × 10 - 24 Am 2

How is a galvanometer converted to an ammeter?

As we know that, a galvanometer is a device that is used to detect small electric current flowing in the circuit and an ammeter is an instrument used to measure the current in a circuit. And ammeter is connected in series with the circuit. It is a low resistance instrument. A Galvanometer is converted into an ammeter by connecting a low resistance S of suitable value in parallel with the galvanometer coil. Here, S = G I g I - I g where, G = Resistance of the Galvanometer

HARD

12th Odisha Board

IMPORTANT

Earn 100

A magnetic field of $4 \times 10^{- 3} \hat{k} tesla$ exerts a force of $(4 \hat{i} + 3 \hat{j}) \times 10^{- 10} newton$ on a particle having a charge of $10^{- 9} coulumb$ in $X - Y$ plane Find the velocity of the particle.

Important Questions on Magnetic Effect of Electric Current

HARD

12th Odisha Board

IMPORTANT

Bureau's Higher Secondary Physics Class 12>Chapter 8 - Magnetic Effect of Electric Current>MODEL QUESTIONS>Q 15

A rectangular coil of

100 turns

and length of

5 cm

and width

4 cm

is suspended in a uniform magnetic field with its plane parallel to the field. A torque of

0.2 N m

acts on the coil when a current of

0.2 A

is maintained in it. Find the magnitude of the magnetic field.

HARD

12th Odisha Board

IMPORTANT

Bureau's Higher Secondary Physics Class 12>Chapter 8 - Magnetic Effect of Electric Current>MODEL QUESTIONS>Q 16

A moving coil galvanometer has a resistance of

30 ohm

, through a current of

1 m A

flows. Find the value of multiplier resistance to be used with the galvanometer to convert it to a voltmeter of the range

2.5 V

HARD

12th Odisha Board

IMPORTANT

Bureau's Higher Secondary Physics Class 12>Chapter 8 - Magnetic Effect of Electric Current>MODEL QUESTIONS>Q 16

A moving coil galvanometer has a resistance of

30 ohm

, through a current of

1 m A

flows. Find the value of multiplier resistance to be used with the galvanometer to convert it to a voltmeter of the range

10 V

HARD

12th Odisha Board

IMPORTANT

Bureau's Higher Secondary Physics Class 12>Chapter 8 - Magnetic Effect of Electric Current>MODEL QUESTIONS>Q 17

A rigidly supported long horizontal wire carries a current

I_{a} = 100 A

. Directly above and parallel to it is a fine wire that carries a current

I_{b} = 20 A

and weigh

0.073 N / m

. How far above the lower wire should this second wire be strung so that it may be supported by magnetic repulsion (assume the diameter of the wire is small)

HARD

12th Odisha Board

IMPORTANT

Bureau's Higher Secondary Physics Class 12>Chapter 8 - Magnetic Effect of Electric Current>MODEL QUESTIONS>Q 18

Two parallel wires separated by

2 m

carry currents of

3 A

each in the opposite directions. Find the magnetic induction for the points between the wires at a distance

50 cm

from each wire.

HARD

12th Odisha Board

IMPORTANT

Bureau's Higher Secondary Physics Class 12>Chapter 8 - Magnetic Effect of Electric Current>MODEL QUESTIONS>Q 19

In the Bohr model of the hydrogen atom, the electron circulates around the nucleus in the path of radius

5.1 \times 10^{11} m

at a frequency of

6.8 \times 10^{15} Hz

. Calculate the magnetic induction B at the centre of the orbit. What is the equivalent dipole moment?

HARD

12th Odisha Board

IMPORTANT

Bureau's Higher Secondary Physics Class 12>Chapter 8 - Magnetic Effect of Electric Current>MODEL QUESTIONS>Q 20

A straight wire segment of length $l$ carries a current $I$ . Show that the field of induction $\vec{B}$ due to it at a distance $R$ from the segment along a perpendicular bisector is given in magnitude by

$B = \frac{μ_{0} I}{2 πR} . \frac{l}{{(l^{2} + 4 R^{2})}^{\frac{1}{2}}}$

Calculate the same for $I = 1 A$ , $l = 1 m$ and $R = 1 m$

MEDIUM

12th Odisha Board

IMPORTANT

Bureau's Higher Secondary Physics Class 12>Chapter 8 - Magnetic Effect of Electric Current>MODEL QUESTIONS>Q 21

25

turn rectangular coil

12 \times 15 cm

with its plane parallel to a flux density of

0.004 W / m^{2}

. Calculate the torque produced in it for the current of

400 mA

when the

12 cm

side is parallel to flux density. Will the torque be different, if the same side is perpendicular to the flux density.

A magnetic field of 4×10-3k ^tesla exerts a force of 4i^+3j^×10-10newton on a particle having a charge of 10-9coulumb in X-Y plane Find the velocity of the particle.

Important Questions on Magnetic Effect of Electric Current

Learn and Prepare for any exam you want !

A magnetic field of $4 \times 10^{- 3} \hat{k} tesla$ exerts a force of $(4 \hat{i} + 3 \hat{j}) \times 10^{- 10} newton$ on a particle having a charge of $10^{- 9} coulumb$ in $X - Y$ plane Find the velocity of the particle.