The deflection in moving coil galvanometer falls from 25 divisions to 5 division when a shunt of $24 \, Omega$ $24 \, \Omega$ is applied. The resistance of galvanometer coil will be:

A. $12 \, Omega$ $12 \, \Omega$
B. $96 \, Omega$ $96 \, \Omega$
C. $48 \, Omega$ $48 \, \Omega$
D. $100 \, Omega$ $100 \, \Omega$

Solution & Explanation

### Related Formula For a shunted galvanometer, the potential difference across the galvanometer is equal to the potential difference across the shunt resistor:

I_g G = (I - I_g) S

### Core Logic Let the current sensitivity per division be

x

. * Initially, without the shunt, the total current

I

gives a full scale deflection of 25 divisions:

I = 25x

Schematic diagrams representing galvanometer before and after shunting for Q44

* When the shunt (

S = 24 \, Omega

) is connected in parallel, the current passing through the galvanometer branch (

I_g

) corresponds to a deflection of 5 divisions:

I_g = 5x

### Step 1: Find Current through the Shunt The current remaining for the shunt branch is:

I - I_g = 25x - 5x = 20x

### Step 2: Equate Potential Drops Applying the parallel branch condition:

5x cdot G = 20x cdot 24

Dividing both sides by

5x

G = 4 times 24 = 96 \, Omega

Therefore, the resistance of the galvanometer coil is

96 \, Omega

. ### Pattern Recognition The deflection is directly proportional to the branch current. A drop from 25 divisions to 5 divisions means only

frac525 = frac15

of the total current flows through the galvanometer, leaving

frac45

to go through the shunt. Hence, the galvanometer resistance must be 4 times the shunt resistance. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Moving Charges and Magnetism

Reference Study Guides

Moving Charges And Magnetism

More Moving Charges and Magnetism Previous-Year Questions — Page 4

Q39 jee_main_2024_29_january_evening Motion of Charged Particle in Magnetic Field

Two particles X and Y having equal charges are being accelerated through the same potential difference. Thereafter they enter normally in a region of uniform magnetic field and describes circular paths of radii

R_1

and

R_2

respectively. The mass ratio of X and Y is:

A. $left(fracR_2R_1right)^2$ $\left(\frac{R_2}{R_1}\right)^2$
B. $left(fracR_1R_2right)^2$ $\left(\frac{R_1}{R_2}\right)^2$
C. $left(fracR_1R_2right)$ $\left(\frac{R_1}{R_2}\right)$
D. $left(fracR_2R_1right)$ $\left(\frac{R_2}{R_1}\right)$

Solution

### Related Formula The radius

R

of the path of a charged particle moving perpendicular to a magnetic field

B

is:

R = fracmvqB = fracpqB

In terms of kinetic energy

K

R = fracsqrt2mKqB

Since the particle is accelerated through potential

V

, kinetic energy

K = qV

R = fracsqrt2mqVqB

R = frac1B sqrtfrac2mVq

### Core Logic For both particles X and Y, the following parameters are the same: * Potential Difference,

V

* Magnetic Field,

B

* Charge,

q

Therefore, we have the proportionality:

R propto sqrtm implies R^2 propto m

### Step 1: Calculate Mass Ratio Using the proportionality relationship:

fracm_1m_2 = left( fracR_1R_2 right)^2

Thus, the mass ratio of X and Y is

left(fracR_1R_2right)^2

. ### Pattern Recognition Shortcut: Whenever charges and potential differences are equal, the radius of orbit in a magnetic field scales as

R propto sqrtm

. Squaring both sides yields

m propto R^2

instantly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Moving Charges and Magnetism

Q54 jee_main_2024_29_january_evening Magnetic Force on a Charge

A charge of

4.0\ mutextC

is moving with a velocity of

4.0 times 10^6text ms^-1

along the positive y-axis under a magnetic field

vecB

of strength

(2hatk)text T

. The force acting on the charge is

xhatitext N

. The value of

x

is ________.

Numerical Answer. Answer: 32 to 32

Solution

### Related Formula The magnetic force on a moving charge is given by the Lorentz force equation:

vecF = q (vecv times vecB)

### Core Logic Given data: * Charge,

q = 4.0\ mutextC = 4.0 times 10^-6text C

* Velocity vector,

vecv = (4.0 times 10^6hatj)text ms^-1

* Magnetic field vector,

vecB = 2hatktext T

### Step 1: Calculate the Force Vector Substitute the values into the force equation:

vecF = (4.0 times 10^-6) left[ (4.0 times 10^6hatj) times (2hatk) right]

vecF = 4.0 times 10^-6 times 8.0 times 10^6 (hatj times hatk)

Since

hatj times hatk = hati

vecF = 32hatitext N

Comparing this to

xhatitext N

, we find:

x = 32

### Pattern Recognition Cross-product check:

hatj times hatk = hati

. The product of the scalar terms

(4.0 times 10^-6) times (4.0 times 10^6) times 2

immediately simplifies to

4 times 4 times 2 = 32

. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Moving Charges and Magnetism

Q34 jee_main_2024_27_jan_morning Lorentz Force

A proton moving with a constant velocity passes through a region of space without any change in its velocity. If

vecE

and

vecB

represent the electric and magnetic fields respectively, then the region of space may have: (A)

E = 0, B = 0

(B)

E = 0, B neq 0

(C)

E neq 0, B = 0

(D)

E neq 0, B neq 0

Choose the most appropriate answer from the options given below:

A. $text(A), (B) and (C) only$ $\text{(A), (B) and (C) only}$
B. $text(A), (C) and (D) only$ $\text{(A), (C) and (D) only}$
C. $text(A), (B) and (D) only$ $\text{(A), (B) and (D) only}$
D. $text(B), (C) and (D) only$ $\text{(B), (C) and (D) only}$

Solution

### Related Formula

vecF_textnet = qvecE + q(vecv times vecB)

### Core Logic For velocity to remain constant, the net force must be zero:

qvecE + q(vecv times vecB) = 0

Let's evaluate the cases: - Case (A): If

E=0

and

B=0

vecF = 0

. (Possible) - Case (B): If

E=0

and

B neq 0

, the magnetic force is zero if

vecv

is parallel or antiparallel to

vecB

(i.e.,

vecv times vecB = 0

). (Possible) - Case (C): If

E neq 0

and

B=0

vecF = qvecE neq 0

, velocity must change. (Not possible) - Case (D): If

E neq 0

and

B neq 0

, the electric and magnetic forces can balance each other perfectly if

qvecE = -q(vecv times vecB)

. (Possible) ### Step 1: Conclusion Hence, statements (A), (B), and (D) represent valid situations where velocity can remain constant. ### Pattern Recognition Velocity filter / velocity selector setups utilize crossed fields where electric fields perfectly balance magnetic components, an archetype of Case (D). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Moving Charges and Magnetism

Q58 jee_main_2024_27_jan_morning Magnetic Field due to Long Straight Wire

Two long, straight wires carry equal currents in opposite directions as shown in the figure. The separation between the wires is

5.0text cm

. The magnitude of the magnetic field at a point

P

midway between the wires is ______

mutextT

. (Given :

mu_0 = 4pi times 10^-7text TcdottextmcdottextA^-1

, and each wire carries a current of

10text A

Magnetic Field due to Long Straight Wire diagram for Q58 - JEE Main 2024 Morning — The diagram maps two vertical wires carrying anti-parallel currents of 10 A separated by 5.0 cm, with central node P denoting the common field contribution site.

Numerical Answer. Answer: 160 to 160

Solution

### Related Formula

B = fracmu_0 i2pi r

### Core Logic Using the right-hand grip rule, both anti-parallel wire systems generate field arrays pointing in the exact same direction at the central midway coordinate. Hence, their field contributions add up directly:

B_textnet = 2 B_1 = 2 left(fracmu_0 i2pi rright) = fracmu_0 ipi r

Where

i = 10text A

, total distance

= 5text cm implies r = 2.5text cm = 2.5 times 10^-2text m

. ### Step 1: Compute numeric value

B_textnet = frac4pi times 10^-7 times 10pi times (2.5 times 10^-2) = frac4 times 10^-62.5 times 10^-2

B_textnet = 1.6 times 10^-4text T = 160 times 10^-6text T = 160\ mutextT

### Pattern Recognition Anti-parallel current pairs generate collaborative field additions inside their spatial boundary zone, rather than destructive structural cancellations. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Moving Charges and Magnetism

Q42 jee_main_2024_29_jan_morning Galvanometer and its Conversions

A galvanometer having coil resistance

10 \, Omega

shows a full scale deflection for a current of

3 mathrm~mA

. For it to measure a current of

8 mathrm~A

, the value of the shunt should be:

A. $3 times 10^-3 \, Omega$ $3 \times 10^{-3} \, \Omega$
B. $4.85 times 10^-3 \, Omega$ $4.85 \times 10^{-3} \, \Omega$
C. $3.75 times 10^-3 \, Omega$ $3.75 \times 10^{-3} \, \Omega$
D. $2.75 times 10^-3 \, Omega$ $2.75 \times 10^{-3} \, \Omega$

Solution

### Related Formula To convert a galvanometer into an ammeter, a small resistance called a shunt (

S

) is connected in parallel with the galvanometer (

G

I_g G = (I - I_g) S implies S = fracI_g GI - I_g

where,

G

= resistance of the galvanometer coil

I_g

= full-scale deflection current of the galvanometer

I

= total current to be measured

S

= shunt resistance ### Core Logic Given values:

G = 10 \, Omega

I_g = 3 mathrm~mA = 3 times 10^-3 mathrm~A

I = 8 mathrm~A

### Step 1: Calculate Shunt Resistance Substituting these values into the shunt formula:

S = frac(3 times 10^-3) times 108 - 0.003

S = frac0.037.997

S approx 3.75 times 10^-3 \, Omega

Therefore, the required value of the shunt resistance is

3.75 times 10^-3 \, Omega

Circuit diagram of conversion of a galvanometer into an ammeter using parallel shunt resistance for Q42

### Pattern Recognition Since

I gg I_g

, the value

I - I_g

in the denominator can be approximated directly as

I

for a very accurate quick shortcut in choice selection:

S approx fracI_g GI = frac0.038 = 3.75 times 10^-3 \, Omega

. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Moving Charges and Magnetism

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The deflection in moving coil galvanometer falls from 25 divisions to 5 division when a shunt of $24 \, Omega$ $24 \, \Omega$ is applied. The resistance of galvanometer coil will be:

Solution & Explanation

Reference Study Guides

More Moving Charges and Magnetism Previous-Year Questions — Page 4

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The deflection in moving coil galvanometer falls from 25 divisions to 5 division when a shunt of 24 \, Omega$24 \, \Omega$ is applied. The resistance of galvanometer coil will be:

Solution & Explanation

Reference Study Guides

More Moving Charges and Magnetism Previous-Year Questions — Page 4

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The deflection in moving coil galvanometer falls from 25 divisions to 5 division when a shunt of $24 \, Omega$ $24 \, \Omega$ is applied. The resistance of galvanometer coil will be: