Logo
Jeeify
MCQEasyJEE 2023Force on Current-Carrying Conductor

JEE Physics 2023 Question with Solution

For a moving coil galvanometer, the deflection in the coil is 0.05 rad0.05 \, \text{rad} when a current of 10 mA10 \, \text{mA} is passed through it. If the torsional constant of the suspension wire is 4.0×10−5 N m/rad4.0 \times 10^{-5} \, \text{N m/rad}, the magnetic field is 0.01 T0.01 \, \text{T}, and the number of turns in the coil is 200200, the area of each turn (in cm2^2) is:

  • A

    2.02.0

  • B

    1.01.0

  • C

    1.51.5

  • D

    0.50.5

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: deflection θ=0.05 rad\theta = 0.05 \, \text{rad}, current i=10×10−3 Ai = 10 \times 10^{-3} \, \text{A}, torsional constant K=4×10−5 N m/radK = 4 \times 10^{-5} \, \text{N m/rad}, magnetic field B=0.01 TB = 0.01 \, \text{T}, and number of turns N=200N = 200.

Find: area of each turn AA.

For a moving coil galvanometer,

τ=Kθ\tau = K\theta

and magnetic torque is

τ=NiAB\tau = NiAB

Comparing the two,

NiAB=KθNiAB = K\theta

Rearranging for AA,

A=KθNiBA = \frac{K\theta}{NiB}

Substituting the given values,

A=(4×10−5)(0.05)200(10×10−3)(0.01)A = \frac{(4 \times 10^{-5})(0.05)}{200(10 \times 10^{-3})(0.01)}

On simplification,

A=10−4 m2=1 cm2A = 10^{-4} \, \text{m}^2 = 1 \, \text{cm}^2

Therefore, the correct option is B.

Direct Torque Comparison

Given: restoring torque is proportional to angular deflection and magnetic torque acts on the current-carrying coil.

Find: the numerical value of area per turn.

At equilibrium, restoring torque equals magnetic torque:

Kθ=NiABK\theta = NiAB

Hence,

A=KθNiBA = \frac{K\theta}{NiB}

Using

K=4×10−5,θ=0.05,N=200,i=10×10−3,B=0.01K = 4 \times 10^{-5}, \quad \theta = 0.05, \quad N = 200, \quad i = 10 \times 10^{-3}, \quad B = 0.01

Now evaluate numerator and denominator separately:

Kθ=(4×10−5)(0.05)=2×10−6K\theta = (4 \times 10^{-5})(0.05) = 2 \times 10^{-6} NiB=200×10−2×10−2=2×10−2NiB = 200 \times 10^{-2} \times 10^{-2} = 2 \times 10^{-2}

Therefore,

A=2×10−62×10−2=10−4 m2A = \frac{2 \times 10^{-6}}{2 \times 10^{-2}} = 10^{-4} \, \text{m}^2

Since

1 m2=104 cm21 \, \text{m}^2 = 10^4 \, \text{cm}^2

we get

10−4 m2=1 cm210^{-4} \, \text{m}^2 = 1 \, \text{cm}^2

So the area of each turn is 1.0 cm21.0 \, \text{cm}^2.

Common mistakes

  • Using A=NiBKθA = \frac{NiB}{K\theta} instead of A=KθNiBA = \frac{K\theta}{NiB}. This inverts the relation and gives an incorrect value. First equate torques correctly and then isolate AA carefully.

  • Not converting 10 mA10 \, \text{mA} into amperes. Taking current as 10 A10 \, \text{A} instead of 10×10−3 A10 \times 10^{-3} \, \text{A} changes the answer by a factor of 10001000. Always convert to SI units before substitution.

  • Forgetting to convert m2\text{m}^2 into cm2\text{cm}^2 at the end. The computed area is 10−4 m210^{-4} \, \text{m}^2, which equals 1 cm21 \, \text{cm}^2, not 10−4 cm210^{-4} \, \text{cm}^2.

Practice more Force on Current-Carrying Conductor questions

Get unlimited AI-adaptive practice, mastery tracking, and an AI tutor that explains every step — free to start.

Start free trialMore on Force on Current-Carrying Conductor

Related questions