Question

1. The magnetic induction field strength at a distance of lcm by the side of an infinitely
long straight conductor carrying current is 2 x 10-T in magnitude. Find the strength
of current through the conductor ?
1) 1A
2) 2 A
3) 3 A​

Answer 1

Answer:

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12th

Physics

Moving Charges and Magnetism

Biot-Savart Law

A long straight wire carryi...

PHYSICS

A long straight wire carrying a current of 30 A is placed in an external uniform magnetic field of induction 4×10

−4

T. The magnetic field is acting parallel to the direction of current. The magnitude of the resultant magnetic induction in tesla at a point 2.0 cm away from the wire is (μ

0

=4π×10

−7

H/m)

MEDIUM

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ANSWER

Magnetic field induction at point P due to current carrying wire is

B

2

=

2πr

μ

0

i

.

2π×0.02

4π×10

−7

×30

B

2

=3×10

−4

T

The direction of B

2

will be perpendicular to B

1

then, the magnitude of the resultant magnetic induction

B=

B

1

2

+B

2

2

=

(4)

2

+(3)

2

×10

−4

=5×10

−4

T

Answer 2

Answer:

The strength of current through the conductor is 1A.

Explanation:

Given the distance from the straight current carrying conductor, $r= 1cm=1\times 10^{-2}m$

The magnetic induction field strength by an infinitely long straight current carrying conductor,  $B= 2 \times 10^{-5}T$

The magnetic field strength by an infinitely long straight current carrying conductor at a radial distance r is given by

$B=\dfrac{\mu_0I}{2\pi r}$

where the permeability of fee space, $\mu_0=4\pi \times 10^{-7}N/A^{2}$ and I is the strength of the current.

Substituting the given values in the formula,

$2\times 10^{-5}=\dfrac{4\pi \times 10^{-7}I}{2\pi \times 1\times 10^{-2}}$

$\implies 2\times 10^{-5}=2\ \times 10^{-5}I}$

$\implies I=1A$

Therefore, the strength of current through the conductor is 1A.

So, the correct answer is option 1.