Question

A length of wire carries a steady current. It is
bent first to form a circular plane coil of one turn.
The same length is now bent more sharply to
give a double loop of smaller radius. The magnetic
field at the centre caused by the same current is
(a) a quarter of its first value
(b) unaltered
(c) four times of its first value
(d) a half of its first value

Answer 1

$\rule{200}{2}$

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Let the radii be r

1

and r

2

respectively.

Since there are two turns of radius r

2

, r

1

=2r

2

Magnetic field B at the center of the coil of radius r

1

B

1

=

2r

1

μ

o

i

=

4r

2

μ

o

i

Magnetic field B at the center of the coil of radius r

2

$$B_2 =2 \times \dfrac{\mu_oi}{2r_2}$$

$$ \therefore \dfrac{B_2}{B_1} = \frac{2 \times \dfrac{\mu_oi}{2r_2}}{\dfrac{\mu_oi}{4r_2}}=4$$

Answer 2

Answer:

Let the radii be r

1

and r

2

respectively.

Since there are two turns of radius r

2

, r

1

=2r

2

Magnetic field B at the center of the coil of radius r

1

B

1

=

2r

1

μ

o

i

=

4r

2

μ

o

i

Magnetic field B at the center of the coil of radius r

2

B_2 =2 \times \dfrac{\mu_oi}{2r_2}B

2

=2×

2r

2

μ

o

i

\therefore \dfrac{B_2}{B_1} = \frac{2 \times \dfrac{\mu_oi}{2r_2}}{\dfrac{\mu_oi}{4r_2}}=4∴

B

1

B

2

=

4r

2

μ

o

i

2×

2r

2

=4