Question

A 2.3 m -long wire carries a current of 8.6 A and is immersed within a uniform magnetic field B⃗ . When this wire lies along the +x axis, a magnetic force F⃗ =(−2.0j)N acts on the wire, and when it lies on the +y axis, the force is F⃗ =(2.0i−4.8k)N.

Answer 1

Answer:

Explanation:

Given that :

The current I = 8.6 A

The length vector L = 2.3 m i

The force vector is = -2.0 j N

When L = 2.3 m i ; the

Force vector F = (2.0 i - 4.8 k) N

Compute the components of the magnetic field as follows:

$F = I(L*B)$

Replacing 8.6 A for I ; -2.0 j N for F & 2.3 m i for L

$-2.0 j N = (8.6 A) (2.3 m i *(B_xi + B_yj+B_zk)$

$-2.0 j N = 19.78B_z \ and \ 0 = 19.78B_y$

$B_z = -0.1011 T \ and \ B_y = 0$

However in y direction ; we have :

$(2.0 i - 4.8 k) = 8,6 A (2.3 mi*(B_xi+B_yj+B_zk)$

$- 2.0 = 19.78 B_z \ and \ -4.8 = 19.78 B_x$

$B_z = -0.1011 T \ and \ B_x = -0.2427T$

Hence, the component of magnetic field is as follows:

$B_x = -.02427T \ ; B_y = 0 T \ ; B_z = - 0.1011 T)$