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Moving Charges and Magnetism

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4 months ago

Moving Charges and Magnetism Physics NCERT Exemplar Solutions Class 12th Chapter Four

A hill is 500 m high. Supplies are to be sent across the hill using a canon that can hurl packets at a speed of 125 m/s over the hill. The canon is located at a distance of 800m from the foot of hill and can be moved on the ground at a speed of 2 m/s; so that its distance from the hill can be adjusted. What is the shortest time in which a packet can reach on the ground across the hill ? Take g =10 m/s².

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alok kumar singh

Contributor-Level 10

This is a Long Answer Type Questions as classified in NCERT Exemplar

Explanation- speed of jackets = 125m/s

Height of hill = 500m

To cross the hill vertical component of velocity should be grater than this value u_y= $\sqrt[]{2 g h}$

$= \sqrt[]{2 * 10 * 500} = 100 m / s$

So u²= u_x²+u_y²

Horizontal component of initial velocity u_x= $\sqrt[]{u^{2} - u_{y}^{2}} = \sqrt[]{125^{2} - 100^{2}} = 75 m / s$

Time taken to reach the top of hill t= $\sqrt[]{\frac{2 h}{g}} = \sqrt[]{\frac{2 * 500}{10}} = 10 s$

Time taken to reach the ground in 10 sec = 75 (10)= 750m

Distance through which the canon has to be moved =800-750=50m

Speed with which canon can move = 2m/s

Time taken canon = 50/2= 25s

Total time t= 25+10+10= 45s

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5 months ago

Moving Charges and Magnetism Class 12th

4.28 A galvanometer coil has a resistance of 15 Ω and the meter shows full scale deflection for a current of 4 mA. How will you convert the meter into an ammeter of range 0 to 6 A?

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Payal Gupta

Contributor-Level 10

4.28 Resistance of the galvanometer coil, G = 15 Ω

Current for which galvanometer shows full deflection, $I_{g}$ = 4 mA = 4 $* 10^{- 3}$ A

Range of ammeter has to be converted from 0 to 6 A, hence I = 6 A

A shunt resistor S is to be connected in parallel with the galvanometer to convert it to an ammeter. The value of S is given as

S = 10 mΩ

Hence, a shunt resistor of 10 mΩ is to be connected to galvanometer to convert it to an ammeter.

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5 months ago

Moving Charges and Magnetism Class 12th

4.27 A galvanometer coil has a resistance of 12 Ω and the meter shows full scale deflection for a current of 3 mA. How will you convert the meter into a voltmeter of range 0 to 18 V?

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Payal Gupta

Contributor-Level 10

4.27 Resistance of the galvanometer coil, G = 12 Ω

Current for which there is full scale deflection, $I_{g}$ = 3 mA = 3 $* 10^{- 3}$ A

Range of voltmeter = 0, to be converted to 18 V, hence V = 18 V

Let there be a resistor R connected in series with the galvanometer to convert it into a voltmeter. R is given as

R = $\frac{V}{I_{g}}$ - G = $\frac{18}{3 * 10^{- 3}}$ - 12 = 5988 Ω

Hence the required value of resistor is 5988 Ω

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5 months ago

Moving Charges and Magnetism Class 12th

4.26 A solenoid 60 cm long and of radius 4.0 cm has 3 layers of windings of 300 turns each. A 2.0 cm long wire of mass 2.5 g lies inside the solenoid (near its centre) normal to its axis; both the wire and the axis of the solenoid are in the horizontal plane. The wire is connected through two leads parallel to the axis of the solenoid to an external battery which supplies a current of 6.0 A in the wire. What value of current (with appropriate sense of circulation) in the windings of the solenoid can support the weight of the wire? g = 9.8 m s^–2.

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Payal Gupta

Contributor-Level 10

4.26 Length of the solenoid, L = 60 cm = 0.6 m

Radius of the solenoid, r = 4.0 cm = 0.04 m

It is given that there are 3 layers of windings of 300 turns each

Hence, total number of turns, n = 900

Length, l = 2 cm = 0.02 m

Mass of the wire, m = 2.5 g = 2.5 $* 10^{- 3}$ kg $* 10^{- 3}$

Current flowing through the wire, I = 6 A

Acceleration due to gravity, g = 9.8 m/ $s^{2}$

We know, magnetic field produced inside the solenoid, B = $\frac{μ_{0} n I}{L}$

where $μ_{0}$ = Permeability of free space = 4 $π * 10^{- 7}$ T m $A^{- 1}$

Magnetic force is given by the relation

F = Bil = $\frac{μ_{0} n I}{L} * i l$

Also the force on the wire is equal to the weight of the wire, F = mg

mg = $\frac{μ_{0} n I i l}{L}$

I = $\frac{m g L}{μ_{0} n i I}$ = $\frac{2.5 * 10^{- 3} * 9.8 * 0.6}{4 π * 10^{- 7} * 900 * 0.02 * 6}$ = 108 A

Hence, the current flowing through the solenoid is 108 A

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4.26 Length of the solenoid, L = 60 cm = 0.6 m

Radius of the solenoid, r = 4.0 cm = 0.04 m

It is given that there are 3 layers of windings of 300 turns each

Hence, total number of turns, n = 900

Length, l = 2 cm = 0.02 m

Mass of the wire, m = 2.5 g = 2.5 $* 10^{- 3}$ kg $* 10^{- 3}$

Current flowing through the wire, I = 6 A

Acceleration due to gravity, g = 9.8 m/ $s^{2}$

We know, magnetic field produced inside the solenoid, B = $\frac{μ_{0} n I}{L}$

where $μ_{0}$ = Permeability of free space = 4 $π * 10^{- 7}$ T m $A^{- 1}$

Magnetic force is given by the relation

F = Bil = $\frac{μ_{0} n I}{L} * i l$

Also the force on the wire is equal to the weight of the wire, F = mg

mg = $\frac{μ_{0} n I i l}{L}$

I = $\frac{m g L}{μ_{0} n i I}$ = $\frac{2.5 * 10^{- 3} * 9.8 * 0.6}{4 π * 10^{- 7} * 900 * 0.02 * 6}$ = 108 A

Hence, the current flowing through the solenoid is 108 A.

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5 months ago

Moving Charges and Magnetism Class 12th

4.25 A circular coil of 20 turns and radius 10 cm is placed in a uniform magnetic field of 0.10 T normal to the plane of the coil. If the current in the coil is 5.0 A, what is the

(a) total torque on the coil,

(b) total force on the coil,

(c) average force on each electron in the coil due to the magnetic field?

(The coil is made of copper wire of cross-sectional area 10^–5 m², and the free electron density in copper is given to be about 1029 m^–3.)

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Payal Gupta

Contributor-Level 10

4.25 Number of turns on the circular coil, n =20

Radius of the coil, r = 10 cm = 0.1 m

Magnetic field strength, B = 0.10 T

Current in the coil, I = 5.0 A

The total torque on the coil is zero because the field is uniform.

Cross-sectional area of the copper coil, A = $10^{- 5}$ $m^{2}$

Number of free electrons per cubic meter of copper, N = $10^{29}$ / $m^{3}$

Charge on the electron, e = 1.6 $* 10^{19}$ C

Magnetic force, F = Be $v_{d}$ where $v_{d}$ is the drift velocity of electrons

$v_{d} = \frac{I}{N e A}$ $μ_{0}$

$H e n c e F$ = $\frac{B e I}{N e A}$ = $\frac{0.1 * 5}{10^{29} * 10^{- 5}}$ = 5 $* 10^{- 25}$ N

The average force on each electron is 5 $* 10^{- 25}$ N

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5 months ago

Moving Charges and Magnetism Class 12th

4.24 A uniform magnetic field of 3000 G is established along the positive z-direction. A rectangular loop of sides 10 cm and 5 cm carries a current of 12 A. What is the torque on the loop in the different cases shown in Fig. 4.28? What is the force on each case? Which case corresponds to stable equilibrium?

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Payal Gupta

Contributor-Level 10

4.24 Magnetic field strength, B = 3000G = 3000 $* 10^{- 4}$ T = 0.3 T

Length of the rectangular loop, l = 10 cm

Width of the rectangular loop, b = 5 cm

Area of the loop, A = l $* b$ = 10 $* 5$ = 50 ${c m}^{2}$ = 50 $* 10^{- 4}$ $m^{2}$

Current in the loop, I = 12 A

Assume that the anti-clockwise direction of the current is positive and vice versa.

Torque, $\overset{?}{τ}$ = $\overset{?}{A}$ $* \overset{?}{B}$

From the given figure, it can be observed that A is normal to the y-z plane and B is directed along z-axis.

$τ$ = 12*( 50 $* 10^{- 4}) \overset{?}{i} * 0.3 \overset{?}{k}$ = $- 1.8 * 10^{- 2} \overset{?}{j}$ Nm

The Torque $1.8 * 10^{- 2}$ is Nm along the negative y-direction.

The force on the loop is zero because the angle between A & B is zero.

This case is similar to case (a). The answer is same as case (a)

Torque, $\overset{?}{τ}$ = I $\overset{?}{A}$ $* \overset{?}{B}$