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NVAEasyJEE 2023Force on Moving Charge

JEE Physics 2023 Question with Solution

A metallic cube of side 15 cm15 \, \text{cm} moving along yy-axis at a uniform velocity of 2 m s−12 \, \text{m s}^{-1}. In a region of uniform magnetic field of magnitude 0.5 T0.5 \, \text{T} directed along zz-axis. In equilibrium the potential difference between the faces of higher and lower potential developed because of the motion through the field will be _____ mV\text{mV}.

Answer

Correct answer:150

Step-by-step solution

Standard Method

Given:

  • Side of the cube, l=15 cm=0.15 ml = 15 \, \text{cm} = 0.15 \, \text{m}
  • Velocity, v⃗=2j^ m/s\vec{v} = 2\hat{j} \, \text{m/s}
  • Magnetic field, B⃗=0.5k^ T\vec{B} = 0.5\hat{k} \, \text{T}

Find: The potential difference developed between the opposite faces of the cube.

For a conductor moving in a magnetic field, the induced potential difference across the appropriate faces is

V=BlvV = Blv

Substituting the given values,

V=(0.5)(0.15)(2)V = (0.5)(0.15)(2) V=0.15 VV = 0.15 \, \text{V}

Converting into millivolts,

V=150 mVV = 150 \, \text{mV}

Therefore, the potential difference developed between the faces is 150 mV150 \, \text{mV}.

Common mistakes

  • Using the wrong formula for magnetic force instead of motional emf. The question asks for potential difference across faces, so use V=BlvV = Blv for the relevant dimension perpendicular to both v⃗\vec{v} and B⃗\vec{B}.

  • Not converting the side length from centimetres to metres. Here 15 cm=0.15 m15 \, \text{cm} = 0.15 \, \text{m}, and failing to convert gives an answer larger by a factor of 100100.

  • Forgetting to convert volts into millivolts at the end. The computed value is 0.15 V0.15 \, \text{V}, which must be written as 150 mV150 \, \text{mV}.

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