A long wire carries a steady current. It is bent into a circle of one turn and the magnetic field at the centre of the coil is B. It is then bent into a circular loop of n turns. The magnetic field at the center of the coil will be

Magnetics PYQ — JEE Main 2004 · Shift 0 · Q142 with Solution

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Correct answer: B

Magnetic field at the center of a single circular loop

For a circular loop of radius $R$ carrying current $I$ , the magnetic field at the center is

$B = \frac{\mu_0 I}{2R}.$

We are told that when the long wire is bent into a circle of one turn, the field at the center is $B$ .

So,

$B = \frac{\mu_0 I}{2R} \quad ...(1)$

where $R$ is the radius of the one-turn circle.

Length of the wire remains constant

Let the total length of the wire be $L$ .

For one turn:

$L = 2\pi R.$

Now the same wire is bent into a circular coil of $n$ turns.

If the radius of each turn is $r$ , then total length is

$L = n(2\pi r).$

Since the wire length is unchanged,

$2\pi R = n(2\pi r)$

$R = nr$

$r = \frac{R}{n}.$

Magnetic field for an $n$ -turn circular coil

Magnetic field at the center of a coil of $n$ turns and radius $r$ is

$B' = \frac{\mu_0 n I}{2r}.$

Substitute $r = \frac{R}{n}$ :

= \frac{\mu_0 n^2 I}{2R}.$$ From equation (1), $$\frac{\mu_0 I}{2R} = B.$$ Hence, $$B' = n^2 B.$$ --- 4. **Check options** - A: $2nB$ — incorrect - B: $n^2B$ — correct - C: $nB$ — incorrect - D: $2n^2B$ — incorrect --- **Final Answer:** $$\boxed{n^2 B}$$ So the correct option is **B**.

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View written solutionFree

Correct answer: B

Magnetic field at the center of a single circular loop

For a circular loop of radius $R$ carrying current $I$ , the magnetic field at the center is

$B = \frac{\mu_0 I}{2R}.$

We are told that when the long wire is bent into a circle of one turn, the field at the center is $B$ .

So,

$B = \frac{\mu_0 I}{2R} \quad ...(1)$

where $R$ is the radius of the one-turn circle.

Length of the wire remains constant

Let the total length of the wire be $L$ .

For one turn:

$L = 2\pi R.$

Now the same wire is bent into a circular coil of $n$ turns.

If the radius of each turn is $r$ , then total length is

$L = n(2\pi r).$

Since the wire length is unchanged,

$2\pi R = n(2\pi r)$

$R = nr$

$r = \frac{R}{n}.$

Magnetic field for an $n$ -turn circular coil

Magnetic field at the center of a coil of $n$ turns and radius $r$ is

$B' = \frac{\mu_0 n I}{2r}.$

Substitute $r = \frac{R}{n}$ :

= \frac{\mu_0 n^2 I}{2R}.$$ From equation (1), $$\frac{\mu_0 I}{2R} = B.$$ Hence, $$B' = n^2 B.$$ --- 4. **Check options** - A: $2nB$ — incorrect - B: $n^2B$ — correct - C: $nB$ — incorrect - D: $2n^2B$ — incorrect --- **Final Answer:** $$\boxed{n^2 B}$$ So the correct option is **B**.

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