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 5

Q56 jee_main_2024_30_january_evening Magnetic Field of a Square Loop

The current of

5 mathrm~A

flows in a square loop of sides

1 mathrm~m

is placed in air. The magnetic field at the centre of the loop is

mathrmXsqrt2 times 10^-7 mathrmT

. The value of

mathrmX

is ________

Numerical Answer. Answer: 40 to 40

Solution

### Related Formula

B = fracmu_0 i4pi d (sintheta_1 + sintheta_2)

### Core Logic For a square loop of side length

a

, the perpendicular distance from the center to any side is

d = fraca2

. The angles subtended by the ends of one side at the center are

theta_1 = theta_2 = 45^circ

. Since there are 4 identical sides contributing to the field in the same direction, the total magnetic field is

4

times the field due to one side. ### Step 1: Calculate Field for One Side

B_1 = fracmu_0 i4pi (a/2) (sin 45^circ + sin 45^circ)

B_1 = frac4pi times 10^-7 times 54pi times 0.5 left( frac1sqrt2 + frac1sqrt2 right)

B_1 = frac10^-7 times 50.5 times frac2sqrt2 = 10 times 10^-7 times sqrt2

### Step 2: Total Magnetic Field

B_texttotal = 4 times B_1 = 4 times 10sqrt2 times 10^-7 mathrm~T

B_texttotal = 40sqrt2 times 10^-7 mathrm~T

Comparing with

Xsqrt2 times 10^-7 mathrmT

, we get

X = 40

. ### Pattern Recognition The magnetic field at the center of an

n

-sided regular polygon carrying current

I

and circumscribed by circle of radius

R

B = fracmu_0 n I2pi R sinleft(fracpinright)

. Or simply use

B = 4 times B_textwire

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

Q39 jee_main_2024_30_jan_morning Magnetic Field at the Center of a Loop

Two insulated circular loop A and B radius 'a' carrying a current of 'I' in the anti clockwise direction as shown in figure. The magnitude of the magnetic induction at the centre will be :

Magnetic Field at the Center of a Loop diagram for Q39 - JEE Main 2024 Morning — Two circular current carrying loops placed orthogonally.

A. $fracsqrt2 mu_0 Ia$ $\frac{\sqrt{2} \mu_0 I}{a}$
B. $fracmu_0 I2 a$ $\frac{\mu_0 I}{2 a}$
C. $fracmu_0 Isqrt2 a$ $\frac{\mu_0 I}{\sqrt{2} a}$
D. $frac2 mu_0 Ia$ $\frac{2 \mu_0 I}{a}$

Solution

### Related Formula

B = fracmu_0 I2R

B_textnet = sqrtB_1^2 + B_2^2 quad (textfor mutually perpendicular fields)

### Core Logic The two loops are mutually perpendicular to each other. The magnetic field at the center due to each loop acts along their respective normal axes, meaning the two resulting magnetic field vectors are orthogonal (

90^circ

apart).

Vector resolution showing orthogonal B fields. — Two circular current carrying loops placed orthogonally.

### Step 1: Calculate Individual Fields Magnetic field at the center due to loop A:

B_A = fracmu_0 I2a

Magnetic field at the center due to loop B:

B_B = fracmu_0 I2a

### Step 2: Resultant Magnetic Field Since

B_A

and

B_B

are perpendicular to each other:

B_textnet = sqrtB_A^2 + B_B^2

B_textnet = sqrtleft(fracmu_0 I2aright)^2 + left(fracmu_0 I2aright)^2

B_textnet = fracmu_0 I2a sqrt2 = fracsqrt2 mu_0 I2a = fracmu_0 Isqrt2 a

### Pattern Recognition When two identical physical sources generate orthogonal vector fields of magnitude

B

, the resultant is simply

sqrt2B

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

Q51 jee_main_2024_30_jan_morning Magnetic Force on Current Carrying Wire

The horizontal component of earth's magnetic field at a place is

3.5 times 10^-5 mathrm~T

. A very long straight conductor carrying current of

sqrt2 mathrm~A

in the direction from South east to North West is placed. The force per unit length experienced by the conductor is

dots dots dots times 10^-6 mathrm~N/m

Numerical Answer. Answer: 35 to 35

Solution

### Related Formula

F = I L B sin theta

fracFL = I B sin theta

### Core Logic Earth's horizontal magnetic field (

B_H

) points strictly from South to North. The conductor is aligned from South-East to North-West. Thus, the angle

theta

between the current flow direction (SE to NW) and Earth's magnetic field (S to N) is exactly

45^circ

. ### Step 1: Compute Force per Unit Length

B_H = 3.5 times 10^-5 mathrm~T

i = sqrt2 mathrm~A

theta = 45^circ

fracFell = i B_H sin theta

fracFell = sqrt2 times (3.5 times 10^-5) times frac1sqrt2

fracFell = 3.5 times 10^-5 mathrm~N/m

fracFell = 35 times 10^-6 mathrm~N/m

### Step 2: Extract Value The required multiplier for

10^-6

is 35. ### Pattern Recognition Geographical alignments often yield clean angles like

45^circ

90^circ

. S-N field intersecting SE-NW current creates a

45^circ

angular intersection, perfectly canceling the

sqrt2

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

Q32 jee_main_2024_31_jan_evening Torque on a Current Loop

A uniform magnetic field of

2 times 10^-3 text T

acts along positive Y-direction. A rectangular loop of sides

20 text cm

and

10 text cm

with current of

5 text A

is Y-Z plane. The current is in anticlockwise sense with reference to negative X axis. Magnitude and direction of the torque is :

A. $2 times 10^-4 text N-m$ $2 \times 10^{-4} \text{ N-m}$ along positive Z-direction
B. $2 times 10^-4 text N-m$ $2 \times 10^{-4} \text{ N-m}$ along negative Z-direction
C. $2 times 10^-4 text N-m$ $2 \times 10^{-4} \text{ N-m}$ along positive X-direction
D. $2 times 10^-4 text N-m$ $2 \times 10^{-4} \text{ N-m}$ along positive Y-direction

Solution

### Related Formula

vecM = ivecA

vectau = vecM times vecB

### Core Logic The area vector

vecA

is perpendicular to the plane of the loop. According to the right-hand rule, an anticlockwise current observed from the negative X-axis means the area vector points in the negative X-direction.

Torque on a Current Loop diagram for Q32 - JEE Main 2024 Evening

### Step 1: Calculating Magnetic Moment Area of the loop,

A = 0.2 text m times 0.1 text m = 0.02 text m^2

Magnetic moment vector:

vecM = i vecA

vecM = 5 times (0.2) times (0.1) (-hati)

vecM = 0.1(-hati) text A m^2

### Step 2: Calculating Torque The magnetic field is

vecB = 2 times 10^-3 hatj text T

. Torque on the loop:

vectau = vecM times vecB

vectau = 0.1(-hati) times (2 times 10^-3 hatj)

vectau = 2 times 10^-4 (-hatk) text N-m

### Step 3: Final Direction The

-hatk

direction corresponds to the negative Z-direction. Magnitude is

2 times 10^-4 text N-m

. ### Pattern Recognition Identify the plane of the loop to get the possible normal vectors. Since the loop is in the Y-Z plane,

vecA

must be

pmhati

. 'Anticlockwise from -X' fixes it as

-hati

. Simply apply cross product

-hati times hatj = -hatk

. ### 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: