Question:

In the cyclotron, as radius of the circular path of the charged particle increases, (\(ω\) = angular velocity, V = linear velocity)

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The wire length is fixed, so n turns have radius R over n.
Updated On: Oct 1, 2026
  • V increases , \(ω\) decreases.
  • V increases , \(ω\) remains constant.
  • only \(ω\) increases , V remains constant.
  • both \(ω\) and V increase.
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The Correct Option is B

Solution and Explanation

Step 1: Radius of the n-turn coil:
Let the single loop have radius R, so the wire length is \(2\pi R\). Forming n turns of radius r gives \(n\cdot2\pi r = 2\pi R\), so \(r = \frac Rn\).

Step 2: Fields at the centre:
One turn: \(B_1 = \frac{\mu_0I}{2R}\). n turns: \(B_n = \frac{n\mu_0I}{2r} = \frac{n\mu_0I\,n}{2R} = \frac{n^2\mu_0I}{2R}\).

Step 3: Ratio:
\(\frac{B_1}{B_n} = \frac{1}{n^2}\), so the ratio is \(1 : n^2\). The factor is \(n^2\), not n, because both the number of turns and the smaller radius raise the field.

Final Answer:
The ratio is \(1 : n^2\), option (C). \[ \boxed{1 : n^2} \]
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