Physics Circle home Student Login

Home›Chapters›Moving Charges and Magnetism PYQs›Ampere's Law, Cylinders, Solenoid & Toroid›Question 17

JEE Main 2025 (29 Jan, Shift 1) · Ampere's Law, Cylinders, Solenoid & Toroid

Ampere's Law, Cylinders, Solenoid & Toroid question from JEE Main 2025 (29 Jan, Shift 1)

Q17JEE Main 2025 · 29 JAN · SHIFT 1Medium

Ampere's Law, Cylinders, Solenoid & Toroid

Consider a long straight wire of a circular cross-section (radius a) carrying a steady current I. The current is uniformly distributed across this cross-section. The distances from the centre of the wire’s cross-section at which the magnetic field [inside the wire, outside the wire] is half of the maximum possible magnetic field, any where due to the wire, will be:

Show answer and solution

Answer: (c)

  1. Maximum field is at the surface: Bmax=μ0I2πaB_{max} = \dfrac{\mu_0 I}{2\pi a}.
  2. Inside: μ0Ir2πa2=Bmax2⇒r=a2\dfrac{\mu_0 I r}{2\pi a^2} = \dfrac{B_{max}}{2} \Rightarrow r = \dfrac{a}{2}.
  3. Outside: μ0I2πr=Bmax2⇒r=2a\dfrac{\mu_0 I}{2\pi r} = \dfrac{B_{max}}{2} \Rightarrow r = 2a.
  4. So the answer is (c) [a/2,2a][a/2, 2a].
Report a mistake

More questions on Ampere's Law, Cylinders, Solenoid & Toroid