Solveeit Logo
Login

Question

Question: A solid cylinderical wire of radius 'R' carries a current T. The magnetic field is 5µT at a point, w...

A solid cylinderical wire of radius 'R' carries a current T. The magnetic field is 5µT at a point, which is '2R' distance away from the axis of wire. Magnetic field at a point which is R/3 distance inside from the surface of the wire is :-

A

103\frac{10}{3}µT

B

203\frac{20}{3}µT

C

53\frac{5}{3}µT

D

403\frac{40}{3}µT

Answer

(2)

Explanation

Solution

  • Formulas for Magnetic Field due to a Solid Cylindrical Wire:

    For a solid cylindrical wire of radius 'R' carrying a current 'I':

    1. Outside the wire (r > R): The magnetic field at a distance 'r' from the axis is given by: Boutside=μ0I2πrB_{outside} = \frac{\mu_0 I}{2 \pi r}

    2. Inside the wire (r < R): Assuming uniform current density, the magnetic field at a distance 'r' from the axis is given by: Binside=μ0Ir2πR2B_{inside} = \frac{\mu_0 I r}{2 \pi R^2}

  • Step 1: Use the given information to find the relationship between μ0I\mu_0 I and R.

    We are given that the magnetic field is 5 µT at a point 2R distance away from the axis. This point is outside the wire (since 2R > R).

    Using the formula for BoutsideB_{outside}:

    5×10−6 T=μ0I2π(2R)5 \times 10^{-6} \text{ T} = \frac{\mu_0 I}{2 \pi (2R)}

    5×10−6=μ0I4πR5 \times 10^{-6} = \frac{\mu_0 I}{4 \pi R}

    From this, we can express μ0IπR\frac{\mu_0 I}{\pi R}:

    μ0IπR=4×(5×10−6)=20×10−6 T\frac{\mu_0 I}{\pi R} = 4 \times (5 \times 10^{-6}) = 20 \times 10^{-6} \text{ T}

  • Step 2: Determine the radial distance for the second point.

    The point is R/3 distance inside from the surface of the wire.

    The distance from the axis of the wire to the surface is R.

    So, the distance from the axis to the point in question (let's call it r') is:

    r′=R−R3=3R−R3=2R3r' = R - \frac{R}{3} = \frac{3R - R}{3} = \frac{2R}{3}

    Since r′=2R3<Rr' = \frac{2R}{3} < R, this point is indeed inside the wire.

  • Step 3: Calculate the magnetic field at the second point.

    Using the formula for BinsideB_{inside}:

    Bnew=μ0Ir′2πR2B_{new} = \frac{\mu_0 I r'}{2 \pi R^2}

    Substitute r′=2R3r' = \frac{2R}{3}:

    Bnew=μ0I(2R/3)2πR2B_{new} = \frac{\mu_0 I (2R/3)}{2 \pi R^2}

    Bnew=μ0I2πR2×2R3B_{new} = \frac{\mu_0 I}{2 \pi R^2} \times \frac{2R}{3}

    Bnew=μ0I3πRB_{new} = \frac{\mu_0 I}{3 \pi R}

  • Step 4: Substitute the value from Step 1 into the expression for BnewB_{new}.

    We found μ0IπR=20×10−6 T\frac{\mu_0 I}{\pi R} = 20 \times 10^{-6} \text{ T}.

    So, Bnew=13×(μ0IπR)B_{new} = \frac{1}{3} \times \left( \frac{\mu_0 I}{\pi R} \right)

    Bnew=13×(20×10−6 T)B_{new} = \frac{1}{3} \times (20 \times 10^{-6} \text{ T})

    Bnew=203×10−6 TB_{new} = \frac{20}{3} \times 10^{-6} \text{ T}

    Bnew=203μTB_{new} = \frac{20}{3} \mu T

The magnetic field at a point which is R/3 distance inside from the surface of the wire is 203\frac{20}{3} µT.