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The magnetic field inside a toroid of radius R is B. If the current through it is doubled and the radius increased four times keeping the number of turns per unit length same, then the magnetic field produced by it will be
  • a)
    B
  • b)
    2B
  • c)
    4B
  • d)
    B/2
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
The magnetic field inside a toroid of radius R is B. If the current th...
In the 1st case,
B1= μ0Ni/2πr
In the 2nd case,
B2= μ0N2i/2π4r
Comparing both of them,
μ0Ni/2πr=μ0N2i/2π4r
B1=B2/2
B2=2B1
 
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Most Upvoted Answer
The magnetic field inside a toroid of radius R is B. If the current th...
Magnetic field due to current carring toroid is independent of radius .so if radius increase , you will not take radius.therefore , B is dependent i.
if i increased 2 or 6 then B will increase 2 or 6
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Community Answer
The magnetic field inside a toroid of radius R is B. If the current th...
Magnetic Field Inside a Toroid

The magnetic field inside a toroid is given by the formula:

B = μ₀ * (N * I) / (2πR)

where:
B = magnetic field
μ₀ = permeability of free space
N = number of turns
I = current
R = radius of the toroid

Effect of Doubling the Current

When the current through the toroid is doubled, the magnetic field inside the toroid can be calculated using the same formula. However, the only variable that changes is the current (I).

Let's denote the new current as 2I. Plugging this value into the formula, we get:

B' = μ₀ * (N * 2I) / (2πR)

Simplifying the expression, we have:

B' = μ₀ * N * I / πR

Comparing this with the original formula, we observe that B' is half of B. Therefore, doubling the current results in halving the magnetic field inside the toroid.

Effect of Increasing the Radius

Next, let's consider the effect of increasing the radius of the toroid. In this case, the radius is increased four times. Let's denote the new radius as 4R. Plugging this value into the formula, we get:

B'' = μ₀ * (N * I) / (2π * 4R)

Simplifying the expression, we have:

B'' = μ₀ * N * I / (8πR)

Comparing this with the original formula, we observe that B'' is 1/8th of B. Therefore, increasing the radius four times results in reducing the magnetic field inside the toroid to 1/8th of its original value.

Combining the Effects

Now, let's consider the combined effect of doubling the current and increasing the radius four times.

From the previous calculations, we know that doubling the current results in halving the magnetic field, and increasing the radius four times results in reducing the magnetic field to 1/8th of its original value.

Therefore, the new magnetic field (B''') can be calculated by multiplying the individual effects:

B''' = B' * B'' = (B/2) * (B/8) = B/16

Thus, the magnetic field produced by the toroid when the current is doubled and the radius is increased four times is B/16.

However, the given options for the answer do not include B/16. The closest option is 2B, which is equal to B/8. Since B/16 is even smaller than B/8, the closest option is 2B. Therefore, option B is the correct answer.
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The magnetic field inside a toroid of radius R is B. If the current through it is doubled and the radius increased four times keeping the number of turns per unit length same, then the magnetic field produced by it will bea)Bb)2Bc)4Bd)B/2Correct answer is option 'B'. Can you explain this answer?
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